hdu 1142 A Walk Through the Forest(最短路)

给一个无向图，求满足要求的最短路条数。

能够从a到达b，当且仅当从a出发到达终点的最短路大于从b出发到达终点的最短路。

以终点为起点，起点为终点跑一遍最短路，求出各个顶点到终点的最短路。然后根据条件进行dfs即可求出答案。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<cstdlib>
#include<cstring>
#include<vector>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef __int64 LI;
typedef unsigned __int64 uLI;
typedef unsigned int uI;
typedef double db;
#define maxn 1005
#define inf 0x3f3f3f3f
struct e
{
    int to,w,next;
}edge[3000005];

int cnt,head[maxn],s,t,n;

inline void add(int u,int v,int w)
{
    edge[cnt].to=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}

bool vis[maxn];
int low[maxn];

void spfa()
{
    int i;
    memset(vis,0,sizeof(vis));
    memset(low,inf,sizeof(low));
    low[2]=0;
    vis[2]=1;
    queue<int> q;
    q.push(2);
    while(!q.empty())
    {
        int u=q.front();q.pop();
        vis[u]=0;
        for(i=head[u]; ~i; i=edge[i].next)
        {
            int v=edge[i].to;
            if(low[v]>low[u]+edge[i].w)
            {
                low[v]=low[u]+edge[i].w;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}

int dp[maxn];

int dfs(int u)
{
    if(dp[u]) return dp[u];
    for(int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        if(low[u]>low[v]) dp[u]+=dfs(v);
    }
    return dp[u];
}
int main()
{
    int n,m;
    while(~scanf("%d",&n)&&n)
    {
        scanf("%d",&m);
        cnt=0;
        memset(head,-1,sizeof(head));
        while(m--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        spfa();
        memset(dp,0,sizeof(dp));
        dp[2]=1;
        printf("%d\n",dfs(1));
    }
    return 0;
}