hdu-1142-A Walk Through the Forest-最短路+记忆化搜索

A Walk Through the Forest

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 61 3 21 4 23 4 31 5 124 2 345 2 247 81 3 11 4 13 7 17 4 17 5 16 7 15 2 16 2 10

 

Sample Output

24

 

题意:问从1到2有多少条路径可以走,规则:对于中途A到B这条路径,如果d(2->B)<d(2->A)则AB这条路径可以走

思路:先求出2到所有点的最短路,然后进行记忆化搜索

#include<stdio.h>#include<string.h>#include<iostream>#include<vector>#include<algorithm>#define inf 10000000000using namespace std;long long w[1005][1005];int v[1005];long long d[1005];int f[1005];int n;void solve(){    memset(v,0,sizeof(v));    for(int i=1 ; i <= n ; i++)        d[i]=inf;    d[2]=0;    for(int i=0 ; i < n ; i++)    {        long long x,m=inf;        for(int y=1 ; y <= n ; y++)            if(!v[y]&&d[y]<=m)            m=d[x=y];        v[x]=1;        for(int y=1; y <=n ; y++)        {            if(d[y]>d[x]+w[x][y])            {                d[y]=d[x]+w[x][y];            }        }    }}int dfs(int pos){    if(pos == 2) return 1;    if(f[pos]) return f[pos];    int xx=0;    for(int i=1 ; i <= n ; i++)    {        if(w[pos][i]>=inf) continue;        if(d[pos] > d[i])        {//            printf("pos=%d i=%d c=%d\n",pos,i,w[pos][i]);            xx+=dfs(i);        }    }    return f[pos]=xx;}int main(){    int m,i,j,a,b;    long long c;    while(scanf("%d",&n)&&n)    {        if(n == 0) break;        scanf("%d",&m);        memset(f,0,sizeof(f));        for(i=0 ; i <= n ; i++)            for(j=0 ; j <= n ; j++)                w[j][i]=w[i][j]=inf;        for(i=0 ; i < m ; i++)        {            scanf("%d%d%lld",&a,&b,&c);            w[a][b]=w[b][a]=c;        }        solve();//        for(int i=1 ; i <= n ; i++)//            printf("%d ",f[i]);        printf("%d\n",dfs(1));    }    return 0;}