Jzoj3934【NOIP2014day2官方数据】寻找道路

先从终点跑一次dijk得到哪几个点是可以连接到终点，让后再跑一次dijk注意不能经过没被标记的点即可

#include<stdio.h>#include<string.h>#include<queue>#include<vector>using namespace std;struct node{ int v,id; };inline bool operator < (node a,node b){ return a.v>b.v; }inline bool min(int& a,int b){ if(a>b){a=b;return 1;} return 0; }vector<int> G[10010],G2[10010];int n,m,d[10010],S,T; bool vis[10010]={0};void Dijk(int s,vector<int> G[10010]){memset(d,127,sizeof d);priority_queue<node> q;d[s]=0; q.push((node){0,s});for(node U;!q.empty();){U=q.top(); q.pop();if(U.v>d[U.id]||vis[U.id]) continue;for(int i=0,u=U.id,v,z=G[u].size();i<z;++i)if(!vis[v=G[u][i]]&&min(d[v],d[u]+1)) q.push((node){d[v],v});}}int main(){scanf("%d%d",&n,&m);for(int x,y,i=0;i<m;++i){scanf("%d%d",&x,&y);G[x].push_back(y);G2[y].push_back(x);}scanf("%d%d",&S,&T);Dijk(T,G2);for(int i=1;i<=n;++i)if(d[i]==0x7f7f7f7f){vis[i]=1;for(int j=0,z=G2[i].size();j<z;++j)vis[G2[i][j]]=1;}Dijk(S,G);if(d[T]==0x7f7f7f7f) puts("-1");else printf("%d\n",d[T]);}