Hdu 1142 A Walk Through the Forest

 题意： 给定一个无向图n*n（m），求从出发点1到终点2中符合以下条件的路径数目。

<条件：若该路径上的任意段（A,B）: 满足B到终点的最短距离比A到终点最短距离要短(即d[A]>d[B]) 则选择B>

思路：Dijkstra+dfs记忆化路径搜索 。

CODE：

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
using namespace std;

const int SIZE = 1001;
const int INF = 0x3f3f3f3f;
int w[SIZE][SIZE];
int v[SIZE], d[SIZE], dp[SIZE];
int n, m;


void init()
{
    memset(w, INF, sizeof(w));
    memset(v, 0, sizeof(v));
    memset(d, 0, sizeof(d));
    memset(dp, 0, sizeof(dp));
}


void Dijkstra(int src)
{
    int i, j;
    for(i = 1; i <= n; i++) d[i] = (i == src)? 0:INF;
    for(i = 1; i <= n; i++)
    {
        int x, m = INF;
        for(int y = 1 ; y <= n; y++) if(!v[y] && m > d[y]) m = d[x=y];
        v[x] = 1;
        for(int y = 1; y <= n; y++) d[y] <?= d[x] + w[x][y];
    }
    return ;
}


int dfs(int v, int e)        //start, end
{
    int i, j;
    int sum = 0;
    if(dp[v]) return dp[v];
    if(v == e) return 1;
    for(i = 1; i <= n; i++)
    {
        if(d[v] > d[i] && w[v][i] != INF)
        {
            int ans = dfs(i, 2);
            sum += ans;
        }
    }
    dp[v] = sum;
    return sum;
}


int main()
{
    while(scanf("%d%d", &n, &m), n )
    {
        init();
        while(m--)
        {
            int u, v, cost;
            scanf("%d%d%d", &u, &v, &cost);
            w[u][v] = w[v][u] = cost;
        }
        Dijkstra(2);
        dfs(1, 2);
        printf("%d\n", dp[1]);
    }
    return 0;
}