Codeforces Gym 100345H Settling the Universe Up Bitset+倒推
2015-09-06 12:50
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题目大意:
给一个只从标号小的点向标号大的点连边的有向图(n<=200)。有如下几种操作:? u v(询问u能否到v)
+ u v (加一条u到v的边)
- u v (删除u到v的边)
以上操作保证u < v,且合法。在操作之前需要给出有多少对(u,v)能够到达,在每次+和-的操作后也要回答上面的问题。
操作数<=100 000,修改操作<=1000
做法:
用bitset维护每个点能到的点对于每一次?操作可以O(1)回答
对于每一次修改操作从后往前推暴力重构bitset并统计答案,时间复杂度O((n+m)∗n)
代码:
4541485 | WHU_FFT | H | Accepted | 852 | 746 | GNU G++ 4.9.2 |
---|
#include <bits/stdc++.h> using namespace std; const int maxn=202; set<int>g[maxn]; set<int>::iterator it; bitset<maxn>b[maxn]; char op[10]; int n; int getans() { int ans=0; for(int i=n;i>0;i--) { b[i].reset(); for(it=g[i].begin();it!=g[i].end();it++) { b[i].set(*it); b[i] |= b[*it]; } ans+=b[i].count(); } return ans; } //倒推暴力重构bitset int main() { freopen("settling.in","r",stdin); freopen("settling.out","w",stdout); int m,t,u,v; scanf("%d%d",&n,&m); for(int i=0;i<m;i++) { scanf("%d%d",&u,&v); g[u].insert(v); } scanf("%d",&t); printf("%d\n",getans()); while(t--) { scanf("%s %d %d",op,&u,&v); if(op[0]=='?') printf("%s\n",(b[u][v])?("YES"):("NO")); else { if(op[0]=='+')g[u].insert(v);else if(op[0]=='-')g[u].erase(v); printf("%d\n",getans()); } } return 0; }
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