A Walk Through the Forest(最短路+搜索)

描述

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.

输入

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.

输出

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

样例输入

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

样例输出

2
4

题意:
起点是1,终点是2,问你从1到2的最短路有几种走法(并不是),真正的意思是:假设 A 和 B 是相连的,当前在 A 处,如果 A 到终点的距离大于 B 到终点的距离,则可以从 A 通往 B 处,问满足这种的条件的数量(路径条数)。

用dis记录2到各个点的最短距离,dfs返回的是从2出发 到这个点的数量(路径条数) 并用dis2记录。
已经存在就直接返回该点的条数,若该点是2就返回1,否则每次加上所有的能到达该点的数量(路径条数)

#include 
using namespace std;
#define inf 0x3f3f3f3f
#define N 1005
int n,m,x,z,y;
int mp[N][N];
int vis[N],dis[N];
int dis2[N];
void f(int kk){
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++){
        dis[i]=mp[kk][i];
    }
    dis[kk]=0;
    vis[kk]=1;
    for(int i=1;i<n;i++){
        int mi=inf,k=0;
        for(int j=1;j<=n;j++){
            if(vis[j]==0&&dis[j]<mi){
                mi=dis[j];
                k=j;
            }
        }
        vis[k]=1;
        for(int j=1;j<=n;j++){
            if(vis[j]==0){
                dis[j]=min(dis[j],dis[k]+mp[k][j]);
            }
        }
    }

}
int dfs(int k){
    if(dis2[k]) return dis2[k];
    if(k==2){
        return 1;
    }
    int sum=0;
    for(int i=1;i<=n;i++){
        if(dis[k]>dis[i]&&mp[k][i]!=inf){
            sum+=dfs(i);
        }
    }
    dis2[k]=sum;
    return dis2[k];
}
int main(){
    while(scanf("%d",&n),n){
        memset(mp,inf,sizeof(mp));
        scanf("%d",&m);
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&x,&y,&z);
            mp[x][y]=z;mp[y][x]=z;
        }
        f(2);

        memset(dis2,0,sizeof(dis2));
        printf("%d\n",dfs(1));
    }
}

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