Hdoj 1428 A Walk Through the Forest 【spfa】+【记忆化搜索】

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6397 Accepted Submission(s): 2348

Problem Description
Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make progress towards his house. He considers taking a path from A to B to be progress if there exists a route from B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take.

Input
Input contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections.

Output
For each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647

Sample Input
5 6
1 3 2
1 4 2
3 4 3
1 5 12
4 2 34
5 2 24
7 8
1 3 1
1 4 1
3 7 1
7 4 1
7 5 1
6 7 1
5 2 1
6 2 1
0

Sample Output
2
4

题意：给你一个无向图，求出从1到2的最短路有几条。
思路：用spfa求出2到各个点之间的距离，然后从1找出有多少条可以到2的点。
代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
const int M = 1007;
const int INF = 0x3f3f3f3f;
using namespace std;

int map[M][M], d[M], ans[M];
int n, m;
bool vis[M];

void spfa(int u){
    for(int i = 1; i <= n; ++ i){
        d[i] = INF; vis[i] = 0;
    }
    vis[u] = 1; d[u] = 0;
    queue<int > q;
    q.push(u);
    while(!q.empty()){
        int temp = q.front(); q.pop(); vis[temp] = 0;
        for(int i = 1; i <= n; ++ i){
            if(d[i] > d[temp]+map[temp][i]){
                d[i] = d[temp]+map[temp][i];
                if(!vis[i]){
                    vis[i] = 1;
                    q.push(i);
                }
            }
        }
    }
}

int dfs(int u){ 
    if(ans[u] != -1) return ans[u];
    if(u == 2) return 1;
    ans[u] = 0;
    for(int i = 1; i <= n; ++ i){
        if(map[u][i] != INF&&d[i] < d[u]){
            ans[u] += dfs(i);
        }
    }
    return ans[u];
}

int main(){
    while(scanf("%d", &n), n){
        scanf("%d", &m);
        for(int i = 0; i <= n; ++ i){
            for(int j = 0; j <= n; ++ j){
                map[i][j] = (i == j?0:INF);
            }
        }
        int a, b, c;
        for(int i = 0; i < m; ++ i){
            scanf("%d%d%d", &a, &b, &c);
            map[a][b] = map[b][a] = min(c, map[a][b]);
        }
        spfa(2);
        memset(ans, -1, sizeof(ans));
        printf("%d\n", dfs(1));
    }
}