[kuangbin带你飞]专题十二基础DP1

A - Max Sum Plus Plus （HDU 1024）

题意：将n个数取m段且不相交，求m段数字和最大值；
dp[i][j]：前i个数字分成j段的最大值。
边界dp[0][0] = 0;
dp[i][j] = max(maxx[j - 1] + a[i], dp[i-1][j] + a[i]);//maxx[j]代表前i个数取j段的最大值。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 5;
int a[maxn], dp[2][maxn], maxx[maxn];int main()
{
//    freopen("in.txt", "r", stdin);std::ios::sync_with_stdio(false);int n, m;while(cin >> m >> n){for(int i = 1; i <= n; i++)cin >> a[i];memset(dp, -0x3f3f3f3f, sizeof(dp));memset(maxx, -0x3f3f3f3f, sizeof(maxx));maxx[0] = 0; dp[0][0] = 0;for(int i = 1; i <= n; i++){int ii = i & 1, tt = 1 - ii;for(int j = 1; j <= min(i, m); j++)dp[ii][j] = max(maxx[j - 1] + a[i], dp[tt][j] + a[i]);for(int j = 1; j <= min(i, m); j++)maxx[j] = max(maxx[j], dp[ii][j]);}cout << maxx[m] << '\n';}return 0;
}

B - Ignatius and the Princess IV（HDU - 1029）

水题

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
map<ll, int> ma;
int main()
{
//    freopen("in.txt", "r", stdin);std::ios::sync_with_stdio(false);int n;while(cin >> n){ll x, ans; ma.clear();for(int i = 1; i <= n; i++){cin >> x; ma[x]++;if(ma[x] >= (n + 1) / 2) ans = x;}cout << ans << endl;}return 0;
}

C - Monkey and Banana （HDU - 1069）

题意：叠箱子，下面的长和宽严格比上面的大，问最高高度；
每个箱子可以旋转所以每种箱子都有6种放法。将它们放入一个结构体数组按长或者宽排序。
对于一个箱子i如果可以放在另一个箱子j上面则可以通过那个箱子转移 dp[i] = max(dp[i],dp[j]+j的长度)；

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<map>
#define endl '\n'
using namespace std;
typedef long long ll;
struct node
{int a, b;int h;int now = 0;
} s[200];
int len;
void add(int a, int b, int c)
{s[len].a = a;s[len].b = b;s[len].now = 0;s[len++].h = c;
}
bool cmp(node a, node b)
{if(a.a != b.a)return a.a < b.a;elsereturn a.b < b.b;
}
int main()
{
//    freopen("in.txt", "r", stdin);std::ios::sync_with_stdio(false);int n; int cas = 1;while(cin >> n){if(n == 0) break;len = 0;for(int i = 1; i <= n; i++){int a, b, c;cin >> a >> b >> c;add(a, b, c); add(b, a, c);add(a, c, b); add(c, a, b);add(c, b, a); add(b, c, a);}sort(s, s + len, cmp);int ans = 0;for(int i = 0; i < len - 1; i++){for(int j = i + 1; j < len; j++){if(s[i].b < s[j].b && s[i].a < s[j].a){s[j].now = max(s[j].now, s[i].now + s[i].h);ans = max(ans, s[j].now + s[j].h);}}}printf("Case %d: maximum height = %d\n", cas++, ans);}return 0;
}

D - Doing Homework （HDU - 1074 ）

题意：有n门课，每门课有截止时间和完成所需的时间，如果超过规定时间完成，每超过一天就会扣1分，问怎样安排做作业的顺序才能使得所扣的分最小
状态压缩+路径保留，每个状态由比它少一门课的状态转移，

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
struct node
{string str;int a, b;
} s[20];
const int maxn = 1 << 15 + 5;
int dp[maxn], pre[maxn], time[maxn];
int main()
{
//    freopen("in.txt", "r", stdin);std::ios::sync_with_stdio(false);int T; cin >> T;while(T--){int n; cin >> n;for(int i = 1; i <= n; i++)cin >> s[i].str >> s[i].a >> s[i].b;memset(dp, 0x3f3f3f3f, sizeof(dp));memset(time, 0x3f3f3f3f, sizeof(time));dp[0] = 0;time[0] = 0;for(int i = 1; i < (1 << n); i++){for(int j = 1; j <= n; j++){int t = 1 << (j - 1);if(!(i & t)) continue;int soc = time[i - t] + s[j].b - s[j].a;if(soc < 0) soc = 0;if(dp[i] >= dp[i - t] + soc){dp[i] = dp[i - t] + soc;time[i] = time[i - t] + s[j].b;pre[i] = j;}}}cout << dp[(1 << n) - 1] << endl;stack<int> st;int now = (1 << n) - 1;while(now){st.push(pre[now]);now -= (1 << (pre[now] - 1));}while(!st.empty()){cout << s[st.top()].str << endl;st.pop();}}return 0;
}

E - Super Jumping! Jumping! Jumping!（ HDU - 1087）

最大的上升子序列+离散化

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 50;
int s[maxn], b[maxn];
map<int, int> pos;
ll dp[maxn];
int main()
{
//    freopen("in.txt", "r", stdin);std::ios::sync_with_stdio(false);int n;while(cin >> n){if(n == 0) break;memset(dp, 0, sizeof(dp));for(int i = 1; i <= n; i++) cin >> s[i], b[i] = s[i];sort(b + 1, b + 1 + n);int len = unique(b + 1, b + 1 + n) - b - 1;for(int i = 1; i <= len; i++)pos[b[i]] = i;ll ans = -1;for(int i = 1; i <= n; i++){int p = pos[s[i]];ll maxx = 0;for(int j = 1; j < p; j++)maxx = max(maxx, dp[j]);dp[p] = max(dp[p], maxx + s[i] * 1ll);ans = max(ans, dp[p]);}cout << ans << endl;}return 0;
}

F - Piggy-Bank （HDU - 1114）

给出一个存钱罐的容量，给出n种硬币的价值p和重量w（注意：每种硬币可无限取）
如果存钱罐能够正好塞满，输出塞满存钱罐需要的最少的硬币价值。
完全背包

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 50;
int dp[maxn];
int main()
{
//    freopen("in.txt", "r", stdin);std::ios::sync_with_stdio(false);int T; cin >> T;while(T--){int e, f; cin >> e >> f;int m = f - e;int n; cin >> n;memset(dp, 0x3f3f3f3f, sizeof(dp));int p, w;dp[0] = 0;for(int i = 1; i <= n; i++){cin >> p >> w;for(int j = 0; j <= m - w; j++)dp[j + w] = min(dp[j + w], dp[j] + p);}if(dp[m] <= 5e8) printf("The minimum amount of money in the piggy-bank is %d.\n", dp[m]);else printf("This is impossible.\n");}return 0;
}

G - 免费馅饼 HDU - 1176

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 50;
struct node
{int x, t;
} s[maxn];
int dp[maxn][11];
bool cmp(node a, node b)
{return a.t < b.t;
}
int now[20];
bool vis[maxn][20];
int main()
{
//    freopen("in.txt", "r", stdin);std::ios::sync_with_stdio(false);int n;while(~scanf("%d", &n)){if(n == 0) break;int T = 0;for(int i = 1; i <= n; i++){scanf("%d%d", &s[i].x, &s[i].t);T = max(T, s[i].t);}sort(s + 1, s + 1 + n, cmp);memset(vis, false, sizeof(vis));memset(dp, 0, sizeof(dp));vis[0][5] = true;int cnt = 1;int ans = 0;for(int i = 1; i <= T; i++){memset(now, 0, sizeof(now));while(s[cnt].t == i && cnt <= n)now[s[cnt++].x]++;for(int j = 0; j <= 10; j++){int maxx = 0;if(j >= 1 && vis[i - 1][j - 1]){maxx = max(maxx, dp[i - 1][j - 1] + now[j]);vis[i][j] = true;}if(vis[i - 1][j + 1]){maxx = max(maxx, dp[i - 1][j + 1] + now[j]);vis[i][j] = true;}if(vis[i - 1][j]){maxx = max(maxx, dp[i - 1][j] + now[j]);vis[i][j] = true;}dp[i][j] = maxx;ans = max(ans, maxx);}}printf("%d\n", ans);}return 0;
}

H - Tickets HDU - 1260

水题

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 2e3 + 50;
int a[maxn], b[maxn];
int dp[maxn];
void func(int t)
{
//  printf("%d\n",t);int h = 8;int m = 0;int s = 0;int ju = 0;for(int i = t; i > 0; i--){s++;if(s >= 60){m++;s -= 60;}if(m >= 60){h++;m -= 60;}if(h == 12) ju = (ju + 1) % 2;if(h > 12)h -= 12;}printf("%02d:%02d:%02d ", h, m, s);if(ju == 1)printf("pm\n");else printf("am\n");
}
int main()
{
//    freopen("in.txt", "r", stdin);int T; cin >> T;while(T--){int n; cin >> n;for(int i = 1; i <= n; i++) cin >> a[i];for(int i = 1; i < n; i++) cin >> b[i];memset(dp, 0x3f3f3f3f, sizeof(dp));dp[1] = a[1];dp[0] = 0;for(int i = 2; i <= n; i++)dp[i] = min(dp[i - 2] + b[i - 1], dp[i - 1] + a[i]);func(dp[n]);}return 0;
}

I - 最少拦截系统 HDU - 1257

贪心+lower_bound

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
vector<int> ve;
int main()
{
//    freopen("in.txt", "r", stdin);int n;while(cin >> n){ve.clear();int ans = 1;int x; cin >> x;ve.push_back(x);vector<int>::iterator it;for(int i = 2; i <= n; i++){cin >> x;if(ve.back() < x){ans++;ve.push_back(x);}else{it = lower_bound(ve.begin(), ve.end(), x);ve.erase(it);ve.insert(it, x);}}cout << ans << endl;}return 0;
}

J - FatMouse’s Speed HDU - 1160

最长上升子序列变形

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
struct node
{int w, s;int pos;
} a[1004];
bool cmp(node a, node b)
{return a.s > b.s;
}
int dp[1005];
int pre[1005];
int main()
{
//    freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);int n = 0;int w, s;while(cin >> w >> s){a[++n].s = s;a[n].w = w;a[n].pos = n;}sort(a + 1, a + 1 + n, cmp);dp[0] = 0;int ans = 0;int ansi;dp[1] = 1;for(int i = 2; i <= n; i++){int maxx = 0;int p;for(int j = 1; a[j].s > a[i].s; j++){if(a[j].w < a[i].w){if(maxx <= dp[j]){maxx = dp[j];p = j;}}}dp[i] = maxx + 1;pre[i] = p;if(ans <= dp[i]){ans = dp[i];ansi = i;}}stack<int> st;cout << ans << endl;while(ans--){st.push(a[ansi].pos);ansi = pre[ansi];}while(!st.empty()){cout << st.top() << endl;st.pop();}return 0;
}

K - Jury Compromise POJ - 1015

难题！
题意：在遥远的国家佛罗布尼亚，嫌犯是否有罪，须由陪审团决定。陪审团是由法官从公众中挑选的。先随机挑选n个人作为陪审团的候选人，然后再从这n个人中选m人组成陪审团。选m人的办法是：
控方和辩方会根据对候选人的喜欢程度，给所有候选人打分，分值从0到20。为了公平起见，法官选出陪审团的原则是：选出的m个人，必须满足辩方总分和控方总分的差的绝对值最小。如果有多种选择方案的辩方总分和控方总分的之差的绝对值相同，那么选辩控双方总分之和最大的方案即可。

开三维的dp数组
dp[i][j][k]代表在第i个人时已经选了j个人分差为k的情况下最大的总分；
因为分差可以是负数所以每个都加上400是它变成正数就能用数组保存了。能够想出来dp的结构的话转移就比较容易了。
dp[i][j][k] = max(dp[i-1][j][k],dp[i-1][j-1][k-差值]+d[i]+p[i]);
另一个难点就是路径保留了。pre数组保存当前状态是通过选择哪个陪审团而来的。
最后尤其要注意看清输出的格式。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
int dp[202][21][805], pre[202][21][805];
int d[300], p[300], cha[300];
int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt","w",stdout);int cas = 1;int n, m;while(~scanf("%d%d", &n, &m)){if(n == 0) break;for(int i = 1; i <= n; i++){scanf("%d%d", &d[i], &p[i]);cha[i] = d[i] - p[i];}//        bianjie
//        memset(pre, 0, sizeof(pre));memset(dp, -0x3f3f3f3f, sizeof(dp));for(int i = 0; i <= n; i++)dp[i][0][400] = 0;for(int i = 1; i <= n; i++){int ch = cha[i];for(int j = min(m, i); j >= 1; j--){for(int k = -400; k <= 400; k++){
//                    dp[j][k+ch] = dp[j-1][k] + d[i]+p[i];if(k + ch + 400 < 0 || k + ch + 400 > 800) continue;dp[i][j][k + ch + 400] = dp[i - 1][j][k + ch + 400];pre[i][j][k + ch + 400] = pre[i - 1][j][k + ch + 400];int t = dp[i - 1][j - 1][k + 400] + d[i] + p[i];if(t > dp[i][j][k + ch + 400]){dp[i][j][k + ch + 400] = t;pre[i][j][k + ch + 400] = i;}}}}printf("Jury #%d\n", cas++);int D = 0, P = 0;int ansi;for(int i = 0; i <= 400; i++){
//            cout<<i-400<<" "<<dp[m][i]<<endl;if(dp[n][m][400 - i] >= 0 || dp[n][m][400 + i] >= 0){ansi = dp[n][m][400 - i] > dp[n][m][400 + i] ? 400 - i : 400 + i;break;}}vector<int> ans;while(m){int t = pre[n][m][ansi];ans.push_back(t);
//            cout<<pre[m][ansi]<<endl;D += d[t]; P += p[t];ansi -= cha[t];m--;n = pre[t - 1][m][ansi];}sort(ans.begin(), ans.end());printf("Best jury has value %d for prosecution and value %d for defence:\n", D, P);for(int i = 0; i < ans.size(); i++)printf(" %d", ans[i]);printf("\n\n");}return 0;
}

L - Common Subsequence POJ - 1458

水题

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
int dp[8000][8000];
char a[10000], b[10000];
int main()
{
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);while(~scanf("%s%s", a + 1, b + 1)){int n = strlen(a + 1);int m = strlen(b + 1);for(int i = 0; i <= n; i++) dp[i][0] = 0;for(int i = 0; i <= m; i++) dp[0][i] = 0;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++){dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);if(a[i] == b[j]) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);}printf("%d\n", dp[n][m]);}return 0;
}

M - Help Jimmy POJ - 1661

按板子高度先排序，不用管高度，因为永远是一个固定的。
dp[i][0]代表到第i块板子的左端需要的时间。dp[i][1]右边。
可以从下往上dp也可以从上往下。
图论的Dijkstra也能写这道题。。。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e4 + 5;
struct node
{int l, r, h;
} s[maxn];
bool cmp(node a, node b)
{return a.h > b.h;
}
int dp[maxn][2];
int main()
{
//    freopen("in.txt", "r", stdin);int T; scanf("%d", &T);while(T--){int n, sx, sy, maxh;scanf("%d%d%d%d", &n, &sx, &sy, &maxh);for(int i = 1; i <= n; i++)scanf("%d%d%d", &s[i].l, &s[i].r, &s[i].h);sort(s + 1, s + 1 + n, cmp);int st = 0;for(int i = 1; i <= n; i++) dp[i][0] = dp[i][1] = 0x3f3f3f3f;for(int i = 1; i <= n; i++){if(s[i].l <= sx && s[i].r >= sx){st = i;dp[i][0] = sx - s[i].l;dp[i][1] = s[i].r - sx;break;}}/*13 8 17 200 10 80 10 134 14 3*/if(st == 0){printf("%d\n", sy);continue;}int ans = 0x3f3f3f3f;for(int i = st; i <= n; i++){bool okl = false, okr = false;for(int j = i + 1; j <= n; j++){
//                cout<<dp[3][0]<<" "<<dp[3][1]<<endl;if(okl && okr) break;if(maxh < s[i].h - s[j].h) break;int L = s[i].l, R = s[i].r;if(!okl)if(s[j].l <= L && s[j].r >= L){okl = true;dp[j][0] = min(dp[j][0], dp[i][0] + L - s[j].l);dp[j][1] = min(dp[j][1], dp[i][0] + s[j].r - L);}if(!okr)if(s[j].l <= R && s[j].r >= R){okr = true;dp[j][0] = min(dp[j][0], dp[i][1] + R - s[j].l);dp[j][1] = min(dp[j][1], dp[i][1] + s[j].r - R);}}if(s[i].h <= maxh){if(!okl)ans = min(ans, dp[i][0] + sy);if(!okr)ans = min(ans, dp[i][1] + sy);}}printf("%d\n", ans);}return 0;
}

N - Longest Ordered Subsequence POJ - 2533

水题

O - Treats for the Cows POJ - 3186

给一个数组a，每次可以取前面的或者后面的，第k次取的a[i]价值为v[i]*k，问总价值最大是多少。
dp[i][j]代表左边取i个右边取j个最大值。
dp[i][j] = max(dp[i - 1][j] + (i + j) * a[i], dp[i][j - 1] + (i + j) * a[n - j + 1]);

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 2e3 + 5;
int dp[maxn][maxn];
int a[maxn];
int main()
{
//    freopen("in.txt", "r", stdin);int n; scanf("%d", &n);for(int i = 1; i <= n; i++) scanf("%d", &a[i]);memset(dp, 0, sizeof(dp));dp[0][0] = 0;for(int i = 1; i <= n; i++)dp[i][0] = dp[i - 1][0] + i * (a[i]);for(int i = 1; i <= n; i++)dp[0][i] = dp[0][i - 1] + i * (a[n - i + 1]);int ans = max(dp[0][n], dp[n][0]);for(int i = 1; i <= n - 1; i++){for(int j = 1; j <= n - i; j++){dp[i][j] = max(dp[i - 1][j] + (i + j) * a[i], dp[i][j - 1] + (i + j) * a[n - j + 1]);if(i + j == n) ans = max(ans, dp[i][j]);}}printf("%d\n", ans);return 0;
}

P - FatMouse and Cheese HDU - 1078

老鼠每次最多垂直或者水平走k步停下来，停下的这个位置只能比上一个停留的位置大，并获取其价值，问最大能得到的价值

将每个点放入结构体按照价值排序。dp[i][j]为（i，j）点最高价值，从小到大遍历转移就行了。这个专题很多题都是类似的思路。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e2 + 5;
struct node
{int x, y, w;
} s[10005];bool cmp(node a, node b)
{return a.w < b.w;
}int ma[maxn][maxn], dp[maxn][maxn], pos[maxn][maxn];
int main()
{
//    freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);int n, k;while(cin >> n >> k){if(n == -1) break;int cnt = 0;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++){cin >> ma[i][j];s[++cnt].w = ma[i][j];s[cnt].x = i;s[cnt].y = j;}sort(s + 1, s + 1 + cnt, cmp);for(int i = 1; i <= cnt; i++)pos[s[i].x][s[i].y] = i;memset(dp, -0x3f3f3f3f, sizeof(dp));dp[1][1] = ma[1][1];int ans = -1;for(int i = 1; i <= cnt; i++){int x = s[i].x;int y = s[i].y;if(dp[x][y] < 0) continue;for(int j = max(1, x - k); j <= min(x + k, n); j++){if(s[pos[j][y]].w > s[i].w) dp[j][y] = max(dp[x][y] + ma[j][y], dp[j][y]);ans = max(ans, dp[j][y]);}for(int j = max(1, y - k); j <= min(y + k, n); j++){if(s[pos[x][j]].w > s[i].w) dp[x][j] = max(dp[x][y] + ma[x][j], dp[x][j]);ans = max(ans, dp[x][j]);}}cout << ans << endl;}return 0;
}

Q - Phalanx HDU - 2859

求矩阵中子矩阵是关于左下角到右上角这条线对称的最大矩阵长度。

看了题解才会。。
dp[i][j]代表以（i,j）为左下角的最大矩阵。

                int x = i - 1, y = j + 1; int tem = 1;while(x >= 0 && y < n && ma[x][j] == ma[i][y])x--, y++, tem++;

这样求出的tem就是i行往上j列往右的对称的长度，最后的dp[i][j] = min(tem, dp[i - 1][j + 1] + 1);

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 5;
char ma[maxn][maxn];
int dp[maxn][maxn];
int main()
{
//    freopen("in.txt", "r", stdin);ios::sync_with_stdio(false);int n;while(cin >> n){if(n == 0) break;for(int i = 0; i < n; i++) cin >> ma[i];int ans = 1;for(int i = 0; i < n; i++){for(int j = 0; j < n; j++){if(i == 0 || j == n - 1) {dp[i][j] = 1; continue;}int x = i - 1, y = j + 1; int tem = 1;while(x >= 0 && y < n && ma[x][j] == ma[i][y])x--, y++, tem++;dp[i][j] = min(tem, dp[i - 1][j + 1] + 1);ans = max(ans, dp[i][j]);}}cout << ans << endl;}return 0;
}

R - Milking Time POJ - 3616

在一个农场里，在长度为N个时间可以挤奶，但只能挤M次，且每挤一次就要休息k分钟；
接下来给m组数据表示挤奶的时间与奶量求最大挤奶量

对于每个时间段的右边界先加上k。然后排序之后就是很简单的dp了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 5;
struct node
{int l, r, w;
} s[maxn];
bool cmp(node a, node b)
{return a.l < b.l;
}
int dp[maxn];
int main()
{
//    freopen("in.txt", "r", stdin);int n, m, k;scanf("%d%d%d", &n, &m, &k);for(int i = 1; i <= m; i++){scanf("%d%d%d", &s[i].l, &s[i].r, &s[i].w);s[i].r += k;}s[0].r = -1;dp[0] = 0;sort(s + 1, s + 1 + m, cmp);int ans = 0;for(int i = 1; i <= m; i++){int maxx = -1;for(int j = 0; j < i; j++){if(s[j].r <= s[i].l && dp[j] > maxx)maxx = dp[j];}dp[i] = maxx + s[i].w;ans = max(ans, dp[i]);}printf("%d\n", ans);return 0;
}

S - Making the Grade POJ - 3666

以前写过