noi 2017 简要题解

回顾noi 2017

DAY1

整数

压位维护序列。用线段树维护一段0后第一个1，一段A-1后第一个<(A - 1)的数
推推转移式子即可

代码犯了无数个错误，并且检查的时候没有仔细的检查出来，仍然借助gdb和对拍才调出来！非常糟糕！

**1. 线段树初始化错误
**2. 打错变量名3处
3. 进位后写成减M，应该是M+1
4. 很多地方一开始想清楚，改的时候没有想清楚，反而改错。比如二分是找最低位非满或非0.修改的时候一定要想清楚，很多时候找到错误反而没有全面的修改，浪费了更多时间
写代码前细节在头脑中不清晰，代码写得太少了！

#include<bits/stdc++.h>
using namespace std;
#define maxn 1002020
#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x)for (register int i = head[x] ; i ; i = e[i].next)
#define pb push_back
#define prev prev_
#define stack stack_
#define mp make_pair
#define fi first
#define se second
#define inf 0x3f3f3f3f
typedef long long ll;
typedef pair<int,int> pr;const ll M = (1 << 30) - 1,K = 30;
int cov[maxn << 2],tag[maxn << 2],val[maxn << 2];
// 1 : 全为M
// 2 ：全为 0
// 0 ：无性质
int n,t1,t2,t3,L;void build(int x,int l,int r){tag[x] = 2 , cov[x] = -1;if ( l == r ) return;int mid = (l + r) >> 1;build(x << 1,l,mid) , build((x << 1) | 1,mid + 1,r);
}
inline void cover(int x,int d){cov[x] = d , val[x] = d;if ( d == M ) tag[x] = 1;else tag[x] = 2;
}
inline void pushdown(int x){int ls = x << 1,rs = (x << 1) | 1;if ( cov[x] != -1 ){cover(ls,cov[x]);cover(rs,cov[x]);cov[x] = -1;}
}
inline void update(int x){int ls = x << 1,rs = (x << 1) | 1;if ( tag[ls] == 1 && tag[rs] == 1 ) tag[x] = 1;else if ( tag[ls] == 2 && tag[rs] == 2 ) tag[x] = 2;else tag[x] = 0;
}
void modify(int x,int l,int r,int L,int R,int d){if ( L > R ) return;if ( L <= l && R >= r ){cover(x,d);return;}int ls = x << 1,rs = (x << 1) | 1,mid = (l + r) >> 1;pushdown(x);if ( L <= mid ) modify(ls,l,mid,L,R,d);if ( R > mid ) modify(rs,mid + 1,r,L,R,d);update(x);
}
void modify(int x,int l,int r,int id,int d){if ( l == r ){val[x] += d;if ( val[x] == M ) tag[x] = 1;else if ( val[x] == 0 ) tag[x] = 2;else tag[x] = 0;return;}int ls = x << 1,rs = (x << 1) | 1,mid = (l + r) >> 1;pushdown(x);if ( id <= mid ) modify(ls,l,mid,id,d);else modify(rs,mid + 1,r,id,d);update(x);
}
int find1(int x,int l,int r,int L,int R){if ( tag[x] == 2 ) return 0;if ( l == r ) return l;int ls = x << 1,rs = (x << 1) | 1,mid = (l + r) >> 1;pushdown(x);if ( L <= l && R >= r ){if ( tag[ls] == 2 ) return find1(rs,mid + 1,r,L,R);return find1(ls,l,mid,L,R);}int p = 0;if ( L <= mid ) p = find1(ls,l,mid,L,R);if ( p ) return p;return find1(rs,mid + 1,r,L,R);
}
int find0(int x,int l,int r,int L,int R){if ( tag[x] == 1 ) return 0;if ( l == r ) return l;int ls = x << 1,rs = (x << 1) | 1,mid = (l + r) >> 1;pushdown(x);if ( L <= l && R >= r ){if ( tag[ls] == 1 ) return find0(rs,mid + 1,r,L,R);return find0(ls,l,mid,L,R);}int p = 0;if ( L <= mid ) p = find0(ls,l,mid,L,R);if ( p ) return p;return find0(rs,mid + 1,r,L,R);
}
int query(int x,int l,int r,int id){if ( l == r ) return val[x];int ls = x << 1,rs = (x << 1) | 1,mid = (l + r) >> 1;pushdown(x);if ( id <= mid ) return query(ls,l,mid,id);return query(rs,mid + 1,r,id);
}
int main(){
//  freopen("input.txt","r",stdin);scanf("%d %d %d %d",&n,&t1,&t2,&t3);L = n + 1000;build(1,0,L);while ( n-- ){int t,a,b,k;scanf("%d",&t);if ( t == 1 ){scanf("%d %d",&a,&b);if ( a < 0 ){a = -a;int x = b / K,y = b % K,t = 0,vx = query(1,0,L,x),vy,delta;ll cur = (ll)a << y;delta = cur & M;if ( vx < delta ) delta -= M + 1 , t++;if ( t || cur >> K ){vy = query(1,0,L,x + 1);int delta2 = t + (cur >> K);if ( vy < delta2 ){int p = find1(1,0,L,x + 2,L);modify(1,0,L,x + 2,p - 1,M);modify(1,0,L,p,-1);delta2 -= M + 1;}modify(1,0,L,x + 1,-delta2);}modify(1,0,L,x,-delta);}else{int x = b / K,y = b % K,t = 0,vx = query(1,0,L,x),vy,delta;ll cur = (ll)a << y;delta = cur & M;if ( vx + delta > M ) delta -= M + 1 , t++;if ( t || cur >> K ){vy = query(1,0,L,x + 1);int delta2 = t + (cur >> K);if ( vy + delta2 > M ){int p = find0(1,0,L,x + 2,L);modify(1,0,L,x + 2,p - 1,0);modify(1,0,L,p,1);delta2 -= M + 1;}modify(1,0,L,x + 1,delta2);}modify(1,0,L,x,delta);}}else{scanf("%d",&a);int x = a / K;int d = query(1,0,L,x);printf("%d\n",(d >> (a % K)) & 1);}}
}

蚯蚓排队

维护hash值，map到1023333的hash表里。
每次分裂只会有k^{2个串被删掉，合并又至多多出k}2个。
O(nk+k^2 * c)

用了两种hash表，还是过不了uoj的extra test，好像还要读入优化，不想卡常了

#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x) for(register int i = head[x] ; i ; i = e[i].next)
#define pb push_back
#define prev prev_
#define stack stack_
#define inf 0x3f3f3f3f
#define maxn 200020
#define N 10000020
#define mp make_pair
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pr;struct hashmap
{int head[10233333] , len[20000010] , cnt[20000010] , next[20000010] , tot;ull val[20000010];inline void insert(int l , ull v){int x = v % 10233333 , i , last = 0;for(i = head[x] ; i ; last = i , i = next[i])if(len[i] == l && val[i] == v)break;if(i) cnt[i] ++ ;else{if(!last) head[x] = ++tot;else next[last] = ++tot;len[tot] = l , val[tot] = v , cnt[tot] ++ ;}}inline void erase(int l , ull v){int x = v % 10233333 , i;for(i = head[x] ; i ; i = next[i])if(len[i] == l && val[i] == v)cnt[i] -- ;}inline int find(int l , ull v){int x = v % 10233333 , i;for(i = head[x] ; i ; i = next[i])if(len[i] == l && val[i] == v)return cnt[i];return 0;}
}rec;
const int MAXK = 50;
const int MAXN = 200005;
const int MAXL = 1e7 + 5;
const int MAXS = 2e7 + 5;
const int HASH = 1e7 + 19;
struct HashTable {int tot, head[HASH];int val[MAXS], nxt[MAXS];ull mask[MAXS];void insert(ull msk, int value) {int p = head[msk % HASH];while (p != 0) {if (mask[p] == msk) {val[p] += value;return;}p = nxt[p];}tot++;mask[tot] = msk;val[tot] = value;nxt[tot] = head[msk % HASH];head[msk % HASH] = tot;}int query(ull msk) {int p = head[msk % HASH];while (p != 0) {if (mask[p] == msk) return val[p];p = nxt[p];}return 0;}
} HT;
const ull base = 97;
const ll mod = 998244353;
int n,m,a[maxn],pre[maxn],post[maxn];
char ch[maxn];
ull pow_[maxn];void init(){pow_[0] = 1;rep(i,1,100) pow_[i] = pow_[i - 1] * base;
}
void merge(int x,int y){int t = 1,id = x; ull cur = 0;while ( t <= 49 ){cur += pow_[t - 1] * a[id];int id2 = y; ull cur2 = cur;rep(j,1,50 - t){cur2 = cur2 * base + a[id2];//  rec.insert(t + j,cur2);HT.insert(cur2,1);id2 = post[id2];if ( !id2 ) break;}id = pre[id] , ++t;if ( !id ) break;}post[x] = y , pre[y] = x;
}
void split(int x){int t = 1,id = x; ull cur = 0;while ( t <= 49 ){cur += pow_[t - 1] * a[id];int id2 = post[x]; ull cur2 = cur;rep(j,1,50 - t){cur2 = cur2 * base + a[id2];//  rec.erase(t + j,cur2);HT.insert(cur2,-1);id2 = post[id2];if ( !id2 ) break;}id = pre[id] , ++t;if ( !id ) break;}  pre[post[x]] = 0 , post[x] = 0;
}
int main(){init();scanf("%d %d",&n,&m);rep(i,1,n){scanf("%d",&a[i]);//  rec.insert(1,a[i]);HT.insert(a[i],1);}rep(i,1,m){int t,x,y,k;scanf("%d",&t);if ( t == 1 ){scanf("%d %d",&x,&y);merge(x,y);}else if ( t == 2 ){scanf("%d",&x);split(x);}else{scanf("%s%d",ch + 1,&k);int l = strlen(ch + 1); ull cur = 0; ll ans = 1;rep(i,1,l){cur = cur * base + (ch[i] - '0');if ( i >= k ){//       ans = ans * rec.find(k,cur) % mod;ans = ans * HT.query(cur) % mod;cur -= pow_[k - 1] * (ch[i - k + 1] - '0');}}printf("%lld\n",ans);}}
}

泳池

直接用大佬的吧
感觉dp还有一种推法，从左往右dp，但是方程没有这么清晰
常系数递推可以用特征多项式优化

DAY2

游戏

枚举x是a还是b型，转换成2-SAT
学到了2-SAT新姿势，输出方案直接比较tarjian后强连通分量的标号，因为标号自然构成逆拓扑序，选标号较小的即可
2-SAT连边的时候，若x1 必须选 y2 ，则 y1 必须选 x2，不能漏连，还要判是否是对同一张图的限制
不要卡常了！平时做题卡常没有意义，把时间花在想和写更多题

#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x) for(register int i = head[x] ; i ; i = e[i].next)
#define pb push_back
#define prev prev_
#define stack stack_
#define inf 0x3f3f3f3f
#define maxn 200020
#define mp make_pair
#define fi first
#define se second
typedef long long ll;
typedef pair<int,int> pr;struct node{int next,to;
}e[maxn * 2],e2[maxn * 2];
struct node2{int x,y;int ax,ay;
}dt[maxn];int head[maxn],cnt,head2[maxn];
int dfn[maxn],ins[maxn],belong[maxn],sz[maxn],low[maxn],st[maxn],tops,dfstime;
int n,m,k,tot,id[maxn],del[maxn],flag;
char ch[maxn],ans[maxn];
pr cur[maxn];inline void adde(int x,int y){if ( x == y ) return;e[++cnt].to = y;e[cnt].next = head[x];head[x] = cnt;
}
void dfs(int x){dfn[x] = low[x] = ++dfstime;ins[x] = 1 , st[++tops] = x;fore(i,x){if ( !dfn[e[i].to] ) dfs(e[i].to) , low[x] = min(low[x],low[e[i].to]);else if ( ins[e[i].to] ) low[x] = min(low[x],low[e[i].to]);}if ( low[x] == dfn[x] ){int y = st[tops--]; ++tot;while ( y != x ){belong[y] = tot , ins[y] = 0;y = st[tops--];}belong[x] = tot , ins[x] = 0;}
}
void init(){rep(i,1,n){if ( ch[i] == 'a' ) del[i] = 1 , cur[i] = mp(i + n,i + n * 2);else if ( ch[i] == 'b' ) del[i + n] = 1 , cur[i] = mp(i,i + n * 2);else if ( ch[i] == 'c' ) del[i + n * 2] = 1 , cur[i] = mp(i + n,i);;}
}
void build(){cnt = tot = dfstime = 0;rep(i,1,n * 3) dfn[i] = low[i] = belong[i] = head[i] = 0;rep(x,1,k){int i = id[x];del[i] = del[i + n] = del[i + n * 2] = 0;if ( ch[i] == 'a' ) del[i] = 1 , cur[i] = mp(i + n,i + n * 2);else if ( ch[i] == 'b' ) del[i + n] = 1 , cur[i] = mp(i,i + n * 2);else if ( ch[i] == 'c' ) del[i + n * 2] = 1 , cur[i] = mp(i + n,i);;}rep(i,1,m){int x = dt[i].x,y = dt[i].y, ax = dt[i].ax , ay = dt[i].ay;if ( del[ax] ) continue;if ( del[ay] ){//  rep(i,0,2) if ( !del[i * n + x] && i * n + x != ax ) adde(ax,i * n + x);adde(ax,cur[x].fi) , adde(ax,cur[x].se);}else{adde(ax,ay);if ( x != y ){// rep(i,0,2) if ( !del[i * n + x] && i * n + x != ax ) //      rep(j,0,2) if ( !del[j * n + y] && j * n + y != ay ) //          adde(j * n + y,i * n + x);if ( ax == cur[x].fi ) {if ( ay == cur[y].fi ) adde(cur[y].se,cur[x].se);else adde(cur[y].fi,cur[x].se);}   else{   if ( ay == cur[y].fi ) adde(cur[y].se,cur[x].fi);else adde(cur[y].fi,cur[x].fi);} }}}
}
bool check(){build();rep(i,1,n * 3) if ( !del[i] && !dfn[i] ) dfs(i);rep(i,1,n){if ( ch[i] == 'a' ){if ( belong[i + n] == belong[i + n * 2] ) return 0;if ( belong[i + n] < belong[i + n * 2] ){ans[i] = 'B';}else ans[i] = 'C';}else if ( ch[i] == 'b' ){if ( belong[i] == belong[i + n * 2] ) return 0;if ( belong[i] < belong[i + n * 2] ){ans[i] = 'A';}else ans[i] = 'C';}else{if ( belong[i + n] == belong[i] ) return 0;if ( belong[i + n] < belong[i] ){ans[i] = 'B';}else ans[i] = 'A';}}return 1;
}
void dfs_x(int x){if ( x == k + 1 ){if ( check() ) flag = 1;return;}ch[id[x]] = 'a';dfs_x(x + 1);if ( flag ) return;ch[id[x]] = 'b';dfs_x(x + 1);
}
int main(){scanf("%d %d",&n,&k);scanf("%s",ch + 1);int cur = 0;rep(i,1,n) if ( ch[i] == 'x' ) id[++cur] = i;scanf("%d",&m);rep(i,1,m){int x,y; char ax[10],ay[10];scanf("%d%s%d%s",&x,ax,&y,ay);dt[i] = (node2){x,y,(ax[0] - 'A') * n + x,(ay[0] - 'A') * n + y};}init();dfs_x(1);if ( !flag ) puts("-1");else{rep(i,1,n) printf("%c",ans[i]);puts("");}
}

蔬菜

这篇博客说得很好
最后可以用并查集维护。
把总量拆成c份，每天的增量是当天坏掉的，这个思路很巧。可以这么拆是因为题目中每天坏掉的是固定的那几个，而卖出后就不会坏了。
读题要看样例解释！！
一开始想错了，以为大的一定先选
不知道为什么一定要倒过来求所有的答案，感觉正着每次选最大的m个也是一样的

update：正着做的确可以。其实原理一样
打错了变量，并查集在x = 0时特判少了，检查代码仍然没有读出来，又借助对拍，非常糟糕！
静下来读代码！！

#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x) for(register int i = head[x] ; i ; i = e[i].next)
#define pb push_back
#define prev prev_
#define stack stack_
#define inf 0x3f3f3f3f
#define maxn 200020
#define mp make_pair
#define fi first
#define se second
typedef long long ll;
typedef pair<int,int> pr;struct node{int a,c,tp,t,x;bool operator < (node cur)const{return a < cur.a;}
};
priority_queue <node> heap;
int n,m,mx,q,p[maxn];
int last[maxn];
int g[maxn * 10],cnt,num[maxn];
ll ans[maxn];
int getlast(int x){if ( x > mx ) return getlast(mx);return x == last[x] ? x : last[x] = getlast(last[x]);
}
int main(){scanf("%d %d %d",&n,&m,&q);rep(i,1,n){int a,x,s,c;scanf("%d %d %d %d",&a,&s,&c,&x);heap.push((node){a + s,1,1,x ? (c - 1) / x + 1 : 100000,x});if ( c > 1 ) heap.push((node){a,c - 1,0,x ? (c - 2) / x + 1 : 100000,x});}rep(i,1,q) scanf("%d",&p[i]) , mx = max(mx,p[i]);rep(i,1,mx) last[i] = i , num[i] = m;while ( heap.size() ){node cur = heap.top(); heap.pop();int a = cur.a , c = cur.c , tp = cur.tp , t = cur.t , x = cur.x , p = 0 , np;if ( tp ){p = getlast(t);if ( p ){g[++cnt] = a;num[p]--;if ( !num[p] ) last[p] = getlast(p - 1);}}else{if ( !x ) p = mx;else p = t , c -= (p - 1) * x;np = getlast(p) , c += (p - np) * x , p = np;while ( p ){if ( !c ) break;int delta = min(num[p],c);rep(i,1,delta) g[++cnt] = a;c -= delta , num[p] -= delta;if ( !num[p] ) last[p] = getlast(p - 1);np = getlast(p - 1) , c += (p - np) * x , p = np;}}}ll sum = 0;rep(i,1,cnt){sum += g[i];ans[(int)ceil((double)i / m)] = sum;}rep(i,1,mx) ans[i] = max(ans[i],ans[i - 1]);rep(i,1,q) printf("%lld\n",ans[p[i]]);
}

分身术

暂时留坑

代码更新中

总结

想题要给自己定时，找性质再确认结论，要把式子推清楚

如果不推清楚式子，只是大概是这样，想了没有任何意义。必须踏踏实实把式子推完整，把细节想清楚！