Codeforces Round #533 (Div. 2) 部分题解A~D

A. Salem and Sticks

题目描述

Salem gave you n n n sticks with integer positive lengths a1,a2,…,an a_1, a_2, \ldots, a_n a1,a2,…,an .

For every stick, you can change its length to any other positive integer length (that is, either shrink or stretch it). The cost of changing the stick's length from a a a to b b b is ∣a−b∣ |a - b| ∣a−b∣ , where ∣x∣ |x| ∣x∣ means the absolute value of x x x .

A stick length ai a_i ai is called almost good for some integer t t t if ∣ai−t∣≤1 |a_i - t| \le 1 ∣ai−t∣≤1 .

Salem asks you to change the lengths of some sticks (possibly all or none), such that all sticks' lengths are almost good for some positive integer t t t and the total cost of changing is minimum possible. The value of t t t is not fixed in advance and you can choose it as any positive integer.

As an answer, print the value of t t t and the minimum cost. If there are multiple optimal choices for t t t , print any of them.

输入输出格式

输入格式：

The first line contains a single integer n n n ( 1≤n≤1000 1 \le n \le 1000 1≤n≤1000 ) — the number of sticks.

The second line contains n n n integers ai a_i ai ( 1≤ai≤100 1 \le a_i \le 100 1≤ai≤100 ) — the lengths of the sticks.

输出格式：

Print the value of t t t and the minimum possible cost. If there are multiple optimal choices for t t t , print any of them.

输入输出样例

输入样例#1： 复制

3
10 1 4

输出样例#1： 复制

3 7

输入样例#2： 复制

5
1 1 2 2 3

输出样例#2： 复制

2 0

直接暴力遍历就行了；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {ll x = 0;char c = getchar();bool f = false;while (!isdigit(c)) {if (c == '-') f = true;c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f ? -x : x;
}ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {if (!b) {x = 1; y = 0; return a;}ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t - a / b * y;return ans;
}
*/int n;
int a[maxn];int main() {//ios::sync_with_stdio(0);cin >> n;for (int i = 1; i <= n; i++)rdint(a[i]);int t = 0;int minn = inf;int pos = 0;for (t = 1; t <= 101; t++) {int sum = 0;for (int i = 1; i <= n; i++) {sum += min(abs(a[i] - t), min(abs(a[i] - (t - 1)), abs(a[i] - (t + 1))));}if (sum < minn) {minn = sum; pos = t;}}cout << pos << ' ' << minn << endl;return 0;
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {ll x = 0;char c = getchar();bool f = false;while (!isdigit(c)) {if (c == '-') f = true;c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f ? -x : x;
}ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {if (!b) {x = 1; y = 0; return a;}ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t - a / b * y;return ans;
}
*/int n, k;
string s;
map<char, int>mp;int main() {//ios::sync_with_stdio(0);cin >> n >> k;cin >> s;int ans = 0;int maxx = -inf;int cnt = 1;int len = 1;for (int i = 1; i < n; i++) {if (s[i] == s[i - 1]) {len++; cnt++;if (i == n - 1) {mp[s[i - 1]] += len / k;}}else {mp[s[i - 1]] += len / k;len = 1; cnt = 1;}}if (n == 1 && k == 1) {cout << 1 << endl; return 0;}for (char ch = 'a'; ch <= 'z'; ch++) {maxx = max(maxx, mp[ch]);}cout << max(0, maxx) << endl;return 0;
}

套路的DP？

设 dp[ i ][ j ]表示前i 个数余数为j的方案数；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {ll x = 0;char c = getchar();bool f = false;while (!isdigit(c)) {if (c == '-') f = true;c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f ? -x : x;
}ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {if (!b) {x = 1; y = 0; return a;}ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t - a / b * y;return ans;
}
*/int n;
int l, r;
ll dp[maxn][3];
int main() {//ios::sync_with_stdio(0);cin >> n >> l >> r;dp[1][0] = r / 3 - l / 3;dp[1][1] = dp[1][0];if (r % 3 > 0)dp[1][1]++;if (l % 3 > 1)dp[1][1]--;dp[1][2] = dp[1][0];if (r % 3 == 2)dp[1][2] ++;if (l % 3 == 0)dp[1][0]++;for (int i = 2; i <= n; i++) {dp[i][0] = ((dp[i - 1][0] * (dp[1][0]) % mod) + (dp[i - 1][1] * dp[1][2]) % mod + (dp[i - 1][2] * dp[1][1]) % mod) % mod;dp[i][1] = ((dp[i - 1][0] * (dp[1][1]) % mod) + (dp[i - 1][1] * dp[1][0]) % mod + (dp[i - 1][2] * dp[1][2]) % mod) % mod;dp[i][2] = ((dp[i - 1][0] * (dp[1][2]) % mod) + (dp[i - 1][1] * dp[1][1]) % mod + (dp[i - 1][2] * dp[1][0]) % mod) % mod;}cout << (ll)dp[n][0] % mod << endl;return 0;
}

多源bfs就行了；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {ll x = 0;char c = getchar();bool f = false;while (!isdigit(c)) {if (c == '-') f = true;c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f ? -x : x;
}ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {if (!b) {x = 1; y = 0; return a;}ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t - a / b * y;return ans;
}
*/int vis[1002][1002];
bool used[1002][1002];
int dis[10][1002][1002];
int n, m;
int p;
int s[10];
vector<pii>vc[10];
int ans[10];
char chp[1003][1003];
int dx[] = { 0,0,1,-1 };
int dy[] = { 1,-1,0,0 };
bool OK(int x, int y) {return x <= n && x >= 1 && y >= 1 && y <= m;
}int main() {ios::sync_with_stdio(0);cin >> n >> m >> p;for (int i = 1; i <= p; i++)cin >> s[i];for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {cin >> chp[i][j];if (!(chp[i][j] == '.') && !(chp[i][j] == '#')) {used[i][j] = chp[i][j] - '0';ans[chp[i][j] - '0']++;vc[chp[i][j]-'0'].push_back(make_pair(i, j));}}}for (int i = 1; i <= p; i++)memset(dis[i], 0x3f, sizeof(dis[i]));while (1) {int fg = 0;for (int i = 1; i <= p; i++) {queue<pii>qu;for (auto it : vc[i]) {int tmpx = it.first;int tmpy = it.second;dis[i][tmpx][tmpy] = 0; qu.push(make_pair(tmpx, tmpy));}vc[i].clear();while (!qu.empty()) {int x = qu.front().first; int y = qu.front().second; qu.pop();if (dis[i][x][y] + 1 > s[i]) {continue;}for (int j = 0; j < 4; j++) {int nx = x + dx[j];int ny = y + dy[j];if (!used[nx][ny] && chp[nx][ny] == '.'&&dis[i][nx][ny] > dis[i][x][y] + 1&&OK(nx,ny)) {dis[i][nx][ny] = dis[i][x][y] + 1;vis[nx][ny] = 1; fg = 1; ans[i]++; used[nx][ny] = 1;qu.push(make_pair(nx, ny)); vc[i].push_back(make_pair(nx, ny));}}}}if (fg == 0)break;}for (int i = 1; i <= p; i++) {cout << ans[i] << ' ';}return 0;
}

转载于:https://www.cnblogs.com/zxyqzy/p/10297314.html