UVA

题意：Jimmy最近打算穿过一个森林，但是他比较傲娇，于是他决定只走一些特殊的道路，他打算只沿着满足如下条件的（A,B）道路走：存在一条从B出发回家的路，比所有从A出发回家的路径都短。你的任务是计算一共有多少条不同的回家路径。其中起点的编号为1，终点的编号为2.

分析：把最短路算出来，迎最短路DAG走，注意记忆搜索DP！

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>#include<queue>#include<vector>const int maxn = 100000;const int inf = 1e9;typedef long long ll;using namespace std;struct HeapNode{    int d,u;    bool operator < (const HeapNode &rhs) const    {        return d > rhs.d;    }};struct Edge{    int fr,to,dist,nex;};struct Dijkstra{    int n,m,cnt;    Edge edges[maxn*2+5];    bool done[maxn+5];    int d[maxn+5],p[maxn+5],head[maxn+5],f[maxn+5];    void init(int n)    {        this->n = n, m = 0, cnt = 0;        for(int i=0; i<=n; i++) head[i] = -1,f[i]=0;        f[2] = 1;    }    void addedge(int fr,int to,int dist)    {        edges[++m] = (Edge)        {            fr,to,dist,head[fr]        };        head[fr] = m;    }    void dijkstra(int s)    {        priority_queue<HeapNode>Q;        for(int i=0; i<=n; i++) d[i] = inf;        d[s] = 0, p[s] =1;        memset(done,0,sizeof(done));        Q.push((HeapNode)        {            0,s        });        while(!Q.empty())        {            HeapNode x = Q.top();            Q.pop();            int u = x.u;            if(done[u]) continue;            done[u] = true;            for(int i=head[u]; i!=-1 ;)            {                Edge& e = edges[i];                i = e.nex;                if(d[e.to]>=d[u]+e.dist)                {                    if(d[e.to]>d[u]+e.dist)                    {                        d[e.to] = d[u] + e.dist;                        Q.push((HeapNode)                        {                            d[e.to],e.to                        });                    }                }            }        }    }    void dfs(int u)    {        if(f[u]) return;        f[u] = 0;        for(int i=head[u]; i!=-1 ;)        {            Edge& e = edges[i];            i = e.nex;            if(d[e.to]<d[u]) dfs(e.to), f[u]+=f[e.to];        }    }} my;int n,m,cnt,p[maxn+5];int main(){    while(scanf("%d",&n)&&n)    {        scanf("%d",&m);        my.init(n);        for(int i=1; i<=m; i++)        {            int x,y,w;            scanf("%d %d %d",&x,&y,&w);            my.addedge(x,y,w), my.addedge(y,x,w);        }        my.dijkstra(2);        my.dfs(1);        printf("%d\n",my.f[1]);    }    return 0;}