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A Walk Through the Forest

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.

InputInput 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.
OutputFor 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


題目大意：我們要從1到達2 問我們有多少種走法。但是我們不能越走越遠，就是我們選擇走這條路，只能距離2這個點越來越近。

思路：我們先dijsktra 點2到所有點的最短路，這樣就能判斷我們走的是不是距離2越來越近。然後dfs搜索，符合題意的路徑，當我們搜索到點2 即可返回。

#include<iostream>  
#include<cstdio>  
#include
 
<cstring>  
#include<ctime>  
#include<cstdlib>  
#include<algorithm>  
#include<cmath>  
#include<string>  
#include<queue>  
#include<vector>  
#include<stack>  
#include<list>  
#include<set>  
#include<map>  
using namespace
 
 std;  
#define P pair<int ,int >  
const int maxn=1000+10;  
const int INF = 0x3f3f3f3f;  
int Lext[maxn],Next[maxn*200],To[maxn*200],Len[maxn*200],dis[maxn],cost[maxn];  
int cnt;  
void add(int u,int v,int w)  
{  
    Next[++cnt]=Lext[u];  
    Lext[u]=cnt;  
    To[cnt]=v;  
    Len[cnt]=w;  
}  
void init()  
{  
    cnt=0;  
    memset(Lext,-1,sizeof(Lext));  
    memset(dis,INF,sizeof(dis));  
    memset(cost,0,sizeof(cost));  
}  
void dij(int st)  
{  
    dis[st]=0;  
    priority_queue<P,vector<P >, greater<P > >q;  
    q.push(P(0,st));  
    while(!q.empty())  
    {  
        P temp=q.top();  
        q.pop();  
        int x=temp.second;  
        for(int i=Lext[x]; i!=-1; i=Next[i])  
        {  
            int y=To[i];  
            int d=Len[i];  
            if(dis[y]>dis[x]+d)  
            {  
                dis[y]=dis[x]+d;  
                q.push(P(dis[y],y));  
            }  
        }  
    }  
}  
int dfs(int st)  
{  
    //cost數組用來存一共有多少走法  
    if(st==2) return 1;  
    if(cost[st]) return cost[st];  
    for(int i=Lext[st]; i!=-1; i=Next[i])  
    {  
        int y=To[i];  
        if(dis[y]<dis[st])  
            cost[st]+=dfs(y);  
    }  
    return cost[st];  
}  
int main()  
{  
    int n,m,u,v,w;  
    while(scanf("%d",&n),n)  
    {  
        init();  
        scanf("%d",&m);  
        for(int i=0; i<m; i++)  
        {  
            scanf("%d %d %d",&u,&v,&w);  
            add(u,v,w);  
            add(v,u,w);  
        }  
        dij(2);  
        cout<<dfs(1)<<endl;  
    }  
}  

PS：摸魚怪的博客分享，歡迎感謝各路大牛的指點~

HDU - 1142 A Walk Through the Forest (最短路)