http://acm.hdu.edu.cn/showproblem.php?pid=2833

大致题意：给定一个无向图，以及悟空和师傅起点与终点，求它们分别从起点到终点的最短路径中经过相同的点的最大个数。

思路：首先dijkstra求出最短路，那么如果有dis[a] + map[a][b] = dis[b]，则边（a,b）一定在最短路径上。根据这一定理可以求出所有最短路径。然后类似于求最长公共子序列求经过的相同点的最大个数。

即若a==b ,dp[a][b] = max(dp[i][j]+1)

否则 dp[a][b] = max(dp[a][j],dp[i][b])，其中i,j分别是a,b的直接前驱节点。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#define LL long long
#define _LL __int64
using namespace std;
const int INF = 1<<28;
const int maxn = 310;

int Map[maxn][maxn];
int n,m;
int s1,t1,s2,t2;
int dis1[maxn],dis2[maxn];
int dp[maxn][maxn];

void init()
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
            Map[i][j] = INF;
    }
}

void dijkstra(int s, int f)
{
    int dis[maxn],vis[maxn];

    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= n; i++)
        dis[i] = Map[s][i];

    vis[s] = 1;
    dis[s] = 0;

    for(int i = 1; i <= n; i++)
    {
        int Min = INF,pos = -1;
        for(int j = 1; j <= n; j++)
        {
            if(!vis[j] && dis[j] < Min)
            {
                Min = dis[j];
                pos = j;
            }
        }
        if(pos == -1) break;
        vis[pos] = 1;
        for(int j = 1; j <= n; j++)
        {
            if(!vis[j] && dis[j] > dis[pos] + Map[pos][j])
                dis[j] = dis[pos] + Map[pos][j];
        }
    }

    if(f == 1)
        memcpy(dis1,dis,sizeof(dis));
    else memcpy(dis2,dis,sizeof(dis));
}

int dfs (int a, int b)
{
    if(a == s1 && b == s2)
        return dp[s1][s2];
    if(dp[a][b] > -1)
        return dp[a][b]; //记忆化

    int v = 0;
    if(a == b)  // dp[a][b] = max(dp[i][j]+1)
    {
    	v++;
        for(int i = 1; i <= n; i++)
        {
            if(dis1[i] + Map[i][a] != dis1[a]) continue;
            for(int j = 1; j <= n; j++)
                if(dis2[j] + Map[j][b] == dis2[b])
                    v = max(v,dfs(i,j)+1);
        }
        return dp[a][b] = v;
    }
	
	//dp[a][b] = max(dp[i][b],dp[a][j]).
    v = 0;
    for(int i = 1; i <= n; i++)
	{
		if(dis1[i] + Map[i][a] == dis1[a])
			v = max(v,dfs(i,b));
	}

	for(int i = 1; i <= n; i++)
	{
		if(dis2[i] + Map[i][b] == dis2[b])
			v = max(v,dfs(a,i));
	}
	return dp[a][b] = v;
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        init();
        if(n == 0 && m == 0) break;
        int u,v,w;

        for(int i = 1; i <= m; i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            if(w < Map[u][v])
                Map[u][v] = Map[v][u] = w;
        }

        scanf("%d %d %d %d",&s1,&t1,&s2,&t2);

        memset(dp,-1,sizeof(dp));
        dp[s1][s2] = 0;
        if(s1 == s2) // 注意起点相同的情况
            dp[s1][s2] = 1;

        dijkstra(s1,1);
        dijkstra(s2,2);


        int ans = dfs(t1,t2);
        printf("%d\n",ans);
    }

    return 0;
}