【记忆化搜索+最短路】HDU 2833 WuKong

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

题意：求2条最短路径的最多公共点数
首先要说的是：求出最短路后，如果dist[u] + map[u][v] = dist[v]，则uv这条边必定在最短路上

#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <set>
//#include <map>
#include <queue>
#include <utility>
#include <iomanip>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
//#include <ctime>
#include <ctype.h>
using namespace std;
#define LL __int64
#define inf 0x3fffffff
#define M 305

int n, map[M][M], d1[M], d2[M], dp[M][M];
bool mark[M];

void init ()
{
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            map[i][j] = inf;
    memset (dp, -1, sizeof(dp));
}

void dijk (int u, int dist[])
{
    memset (mark, false, sizeof(mark));
    int mins, i, j, v;
    for (i = 1; i <= n; i++)
        dist[i] = map[u][i];
    dist[u] = 0;
    mark[u] = true;
    while (1)
    {
        mins = inf;
        for (j = 1; j <= n; j++)
            if (!mark[j] && dist[j] < mins)
                mins = dist[j], v = j;
        if (mins == inf)
            break;
        mark[v] = true;
        for (j = 1; j <= n; j++)
            if (!mark[j] && dist[v] + map[v][j] < dist[j])
                dist[j] = dist[v] + map[v][j];
    }
}

int dfs (int a, int b)
{
    if (dp[a][b] > -1)		//记忆
        return dp[a][b];
    int i, j, v = 0;    //v是重要的临时值
    if (a == b)				//a==b可以一起往前走一步
    {
        v++;
        for (i = 1; i <= n; i++)		//枚举第一条最短路的可以到达a的前一个点
        {
            if (d1[i] + map[i][a] != d1[a])		//ia不是最短路上的边
                continue;
            for (j = 1; j <= n; j++)	//枚举第二条最短路的可以到达b的前一个点
                if (d2[j] + map[j][b] == d2[b])
                    v = max (v, dfs (i, j)+1);
        }
    }
    for (i = 1; i <= n; i++)		//a往前走一步
        if (d1[i] + map[i][a] == d1[a])
            v = max (v, dfs (i, b));
    for (i = 1; i <= n; i++)		//b往前走一步
        if (d2[i] + map[i][b] == d2[b])
            v = max (v, dfs (a, i));
    dp[a][b] = v;
    return v;
}

int main()
{
    int m, u, v, w, A, B, C, D;
    while (scanf ("%d%d", &n, &m), (n||m))
    {
        init ();
        while (m--)
        {
            scanf ("%d%d%d", &u, &v, &w);
            if (w < map[u][v])
                map[u][v] = map[v][u] = w;
        }
        scanf ("%d%d%d%d", &A, &B, &C, &D);
        dp[A][C] = 0;
        if (A == C)			//dfs重要返回条件
            dp[A][C] = 1;
        dijk (A, d1);
        dijk (C, d2);
        printf ("%d\n", dfs (B, D));	//dfs(A, C)会超时……要从终点走回起点
    }
    return 0;
}