The King’s Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem Description
In the Kingdom of Silence, the king has a new problem. There are N cities in the kingdom and there are M directional roads between the cities. That means that if there is a road from u to v, you can only go from city u to city v, but can’t go from city v to city u. In order to rule his kingdom more effectively, the king want to divide his kingdom into several states, and each city must belong to exactly one state. What’s more, for each pair of city (u, v), if there is one way to go from u to v and go from v to u, (u, v) have to belong to a same state. And the king must insure that in each state we can ether go from u to v or go from v to u between every pair of cities (u, v) without passing any city which belongs to other state.
Now the king asks for your help, he wants to know the least number of states he have to divide the kingdom into.
Now the king asks for your help, he wants to know the least number of states he have to divide the kingdom into.
Input
The first line contains a single integer T, the number of test cases. And then followed T cases.
The first line for each case contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the number of cities and roads in the kingdom. The next m lines each contains two integers u and v (1 <= u, v <= n), indicating that there is a road going from city u to city v.
The first line for each case contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the number of cities and roads in the kingdom. The next m lines each contains two integers u and v (1 <= u, v <= n), indicating that there is a road going from city u to city v.
Output
The output should contain T lines. For each test case you should just output an integer which is the least number of states the king have to divide into.
Sample Input
1
3 2
1 2
1 3
Sample Output
2
Source
题意：给你n个点，m条边，可以将一个单联通分量缩成一个点，最少能分成几个点；
思路：先将强连通分量缩点，强连通肯定是可以合并成一个点，然后求无环DAG图的最小路径覆盖即可；
#include<iostream> #include<cstdio> #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> #include<bitset> using namespace std; #define LL unsigned long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=5e3+10,M=1e5+10,inf=1e9+10; const LL INF=1e18+10,mod=2147493647; struct is { int u,v; int next; }edge[M]; int head[N]; int belong[N]; int dfn[N]; int low[N]; int stackk[N<<1]; int instack[N]; int number[N]; int n,m,jiedge,lu,bel,top; void update(int u,int v) { jiedge++; edge[jiedge].u=u; edge[jiedge].v=v; edge[jiedge].next=head[u]; head[u]=jiedge; } void dfs(int x) { dfn[x]=low[x]=++lu; stackk[++top]=x; instack[x]=1; for(int i=head[x];i;i=edge[i].next) { if(!dfn[edge[i].v]) { dfs(edge[i].v); low[x]=min(low[x],low[edge[i].v]); } else if(instack[edge[i].v]) low[x]=min(low[x],dfn[edge[i].v]); } if(low[x]==dfn[x]) { int sum=0; bel++; int ne; do { sum++; ne=stackk[top--]; belong[ne]=bel; instack[ne]=0; }while(x!=ne); number[bel]=sum; } } void tarjan() { memset(dfn,0,sizeof(dfn)); bel=lu=top=0; for(int i=1;i<=n;i++) if(!dfn[i]) dfs(i); } void init() { memset(head,0,sizeof(head)); jiedge=0; } vector<int> g[N]; int cy[N]; bool vis[N]; bool dfs1(int u){ for(int i=0; i<g[u].size(); ++i){ int v = g[u][i]; if(vis[v]) continue; vis[v] = true; if(cy[v]==-1 || dfs1(cy[v])){ cy[v] = u; return true; } } return false; } int solve(int n){ int ret = 0; memset(cy, -1, sizeof(cy)); for(int i=1;i<=n;++i){ memset(vis, 0, sizeof(vis)); ret += dfs1(i); } return n - ret; } int main() { int T; scanf("%d",&T); while(T--) { init(); scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) g[i].clear(); for(int i=1;i<=m;i++) { int u,v; scanf("%d%d",&u,&v); update(u,v); } tarjan(); for(int i=1;i<=jiedge;i++) { if(belong[edge[i].v]!=belong[edge[i].u]) { g[belong[edge[i].u]].push_back(belong[edge[i].v]); } } int ans=solve(bel); printf("%d ",ans); } return 0; }