HDU 3313 Key Vertex（BFS+BFS） 求S点到T点路径的关键点

Key Vertex

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1349    Accepted Submission(s): 307

Problem Description

You need walking from vertex S to vertex T in a graph. If you remove one vertex which stops you from walking from S to T, that vertex we call as key vertex. Now you are given a directed graph, S and T, and you should tell us how many key vertexes are there in the graph.
Please notice that S and T are key vertexes and if S cannot walking to T by the directed edge in the initial graph then all vertexes becomes to key vertexes.

 

Input

The input consists of multiply test cases. The first line of each test case contains two integers, n(0 <= n <= 100000), m(0 <= m <= 300000), which are the number of vertexes and the number of edge. Each of the next m lines consists of two integers, u, v(0 <= u, v < n; u != v), indicating there exists an edge from vertex u to vertex v. There might be multiple edges but no loops. The last line of each test case contains two integers, S, T(0 <= S, T < n, S != T).

 

Output

Output the number of key vertexes in a single line for each test case.

 

Sample Input

6 60 11 21 32 43 44 50 5

 

Sample Output

4

 

Author

momodi

 

Source

HDOJ Monthly Contest – 2010.02.06

 

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题意：给出一个有向无环图，问从S点到T点有多少个关键点（即每条路都必过的点），如果没有路从S到T，那么所有的点都是关键点。

解题：因为要求关键点，那么我们可以求一条从S到T的最短路，那么所有的关键点必然在这一条最短路上，我们把这条最短路上所有的点按经过的倒至顺序从0都标一个号，其他的点都标为-1，每一次第一个加入队列里面的都是必须点。再找的过程中更新必须点的指向。

参考详解链接

#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define MAXN 100005#define MAXM 300005struct EDG{    int to,next;}edg[MAXM];int eid,head[MAXN];void init(){    eid=0;    memset(head,-1,sizeof(head));}void addedg(int u,int v){    edg[eid].to=v; edg[eid].next=head[u]; head[u]=eid++;}int vist[MAXN],flagMark[MAXN];bool bfs1(int s,int t){    memset(vist,-1,sizeof(vist));    vist[s]=-2;    queue<int>q;    q.push(s);    while(!q.empty())    {        int u=q.front();  q.pop();        for(int i=head[u]; i!=-1; i=edg[i].next)        {            int v=edg[i].to;            if(vist[v]==-1)                vist[v]=u,q.push(v);            if(v==t)            {                memset(flagMark,-1,sizeof(flagMark));                int mark=0;                while(v!=-2)                {                    flagMark[v]=mark;                    mark++;                    v=vist[v];                }                return 1;            }        }    }    return 0;}int bfs2(int s,int t,int n){    bool flag=bfs1(s,t);    if(flag==0)        return n;    queue<int>q;    memset(vist,0,sizeof(vist));    int mark=flagMark[s];    int mustnode=s;    int ans=2;    vist[s]=1;    q.push(s);    while(!q.empty())    {        while(!q.empty())        {            int u=q.front(); q.pop();            for(int i=head[u]; i!=-1; i=edg[i].next)            {                int v=edg[i].to;                if(flagMark[v]==-1)                {                    if(vist[v]==0)                        vist[v]=1,q.push(v);                }                else if(flagMark[v]<mark) //更新必须点的指向                {                    mark=flagMark[v];                    mustnode=v;                    if(mustnode==t)                        return ans;                }            }        }        ans++;        q.push(mustnode);    }    return ans;}int main(){    int n,m,u,v;    int s,t;    while(scanf("%d%d",&n,&m)>0)    {        init();        while(m--)        {            scanf("%d%d",&u,&v);            addedg(u,v);        }        scanf("%d%d",&s,&t);        printf("%d\n",bfs2(s,t,n));    }}