hoj1142

最短路+dfs，dfs的时候要记忆化，并记得判断能否从该点来。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define INF 0x03F3F3F3F
const int N = 1000;

int path[N], vis[N];
int n, m;
int cost[N][N], lowcost[N], dp[N];

void input()
{
    scanf("%d", &m);
    memset(cost, -1, sizeof(cost));
    for (int i = 0; i < m; i++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        a--;
        b--;
        cost[a][b] = c;
        cost[b][a] = c;
    }
}

void Dijkstra(int beg)
{
    int i, j, min;
    memset(vis, 0, sizeof(vis));
    vis[beg] = 1;
    for (i=0; i<n; i++)
    {
        lowcost[i] = INF; path[i] = beg;
    }
    lowcost[beg] = 0;
    path[beg] = -1; // 树根的标记
    int pre = beg;
    for (i=1; i<n; i++)
    {
        min = INF;
        for (j=0; j<n; j++)
        // 下面的加法可能导致溢出,INF不能取太大
            if (vis[j]==0 && cost[pre][j] != -1 && lowcost[pre]+cost[pre][j]<lowcost[j])
            {
                lowcost[j] = lowcost[pre] + cost[pre][j];
                path[j] = pre;
            }
        for (j=0; j<n; j++)
            if (vis[j] == 0 && lowcost[j] < min)
            {
                min = lowcost[j]; pre = j;
            }
        vis[pre] = 1;
        //cout << lowcost[pre] << endl;
    }
}

int dfs(int a)
{
    if (dp[a] != -1)
        return dp[a];
    int ans = 0;
    for (int i = 0; i < n; i++)
        if (cost[i][a] != -1 && lowcost[i] > lowcost[a])
            ans += dfs(i);
    dp[a] = ans;
    return dp[a];
}

int main()
{
//    freopen("D:\\t.txt", "r", stdin);
    while (scanf("%d", &n) != EOF && n != 0)
    {
        input();
        Dijkstra(1);
        memset(dp, -1, sizeof(dp));
        dp[0] = 1;
        printf("%d\n", dfs(1));
    }
    return 0;
}