poj 2662 A Walk Through the Forest

A Walk Through the Forest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2385 Accepted: 880

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.

思路： 从2开始的最短路 求 dist[] 表示离家的最近距离；在从1 开始搜，搜的过程优先使dist[i] 比较大的优先更新相邻的点的dp;

dp[i]= sum { dp[j] }  (pos[i][j] !=-1 && dist[j] > dist[i] );

#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<list>#include<map>#include<set>using namespace std;int n,m;const int INF=0xfffffff;const int N=1010;int dist[N],dp[N],pos[N][N];bool vis[N];struct node{    int i,dist;    node(int i=0,int dist=0):i(i),dist(dist){}    bool operator < (const node a)const{        return dist < a.dist;    }};vector<int>e[N];void init(){    for(int i=1;i<=n;i++)        e[i].clear();    memset(pos,-1,sizeof(pos));    scanf("%d",&m);    int a,b,c;    for(int i=0;i<m;i++){        scanf("%d%d%d",&a,&b,&c);        if(pos[a][b]==-1){            e[a].push_back(b);            e[b].push_back(a);            pos[a][b]=pos[b][a]=c;        }        else{            pos[a][b]=min(pos[a][b],c);            pos[b][a]=pos[a][b];        }    }}void deal_min(int &a,int b){    if(b==-1) return ;    if(a==-1) a=b;    else a=min(a,b);}void deal_dist(){    queue<int>que;    memset(vis,0,sizeof(vis));    memset(dist,-1,sizeof(dist));    que.push(2);vis[2]=1;dist[2]=0;    while(!que.empty()){        int i=que.front();que.pop();        vis[i]=0;        if(i==1) continue;        vector<int>::iterator it;        for(it=e[i].begin();it!=e[i].end();it++){            int tp=dist[*it];            deal_min(dist[*it],dist[i]+pos[i][*it]);            if(tp!=dist[*it] && vis[*it]==0){                vis[*it]=1;                que.push(*it);            }        }    }}//void deal_dp(){   //使用优先队列//    memset(dp,0,sizeof(dp));//    memset(vis,0,sizeof(vis));////    priority_queue<node>que;//    que.push(node(1,dist[1]));//    vis[1]=1;//    dp[1]=1;////    while(!que.empty()){//        node cur=que.top();que.pop();//        int i=cur.i;//        if(i==2) continue;//        vector<int>::iterator it;//        for(it=e[i].begin();it!=e[i].end();it++){//            if(dist[*it]<dist[i])//                dp[*it]+=dp[i];//            if(vis[*it]==0){//                que.push(node(*it,dist[*it]));//                vis[*it]=1;//            }//        }//    }//    printf("%d\n",dp[2]);////}void deal_dp(){    memset(dp,0,sizeof(dp));    memset(vis,0,sizeof(vis));    queue<int>que;    que.push(1);    vis[1]=1;dp[1]=1;    while(!que.empty()){        int i=que.front();que.pop();        if(i==2) continue; //如果没有这句下面的清零要标注 i != 2;        vector<int>::iterator it;        for(it=e[i].begin();it!=e[i].end();it++){            if(dist[*it]<dist[i])                dp[*it]+=dp[i];            if(vis[*it]==0&& dist[*it]<dist[i]){                que.push(*it);                vis[*it]=1;            }        }        vis[i]=0;dp[i]=0; //取消标记 dp清零，避免重复计算；    }    printf("%d\n",dp[2]);}int main(){//freopen("in.in","r",stdin);    while(~scanf("%d",&n)&&n){        init();        deal_dist();        deal_dp();    }return 0;}//5 6//1 3 3//1 4 1//4 5 1//5 3 1//5 2 5//3 2 4////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