hdu 1142(迪杰斯特拉+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): 7330    Accepted Submission(s): 2687

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

 

从1号点走到二号点，下一步到2号点永远要比上一步离2号点近（比如第一个测试样例中2-3 为37 1-2为36，所以肯定不会走3号点）。求符合这样的路径条数。

先以2号点做迪杰斯特拉，求出它到每个点的最小距离，然后记忆化搜索。

做了这个题，我推荐还可以做下 hdu 1428

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int N = 1005;
const int INF = 99999999;
int graph[N][N];
int n,m;
bool vis[N];
int low[N];
int dp[N];
void dijkstra(int n,int pos)
{
    memset(vis,0,sizeof(vis));
    vis[pos]=1;
    for(int i=1; i<=n; i++) low[i]=graph[pos][i];
    low[pos] = 0;
    for(int i=1; i<n; i++)
    {
        int Min=INF;
        for(int j=1; j<=n; j++)
            if(vis[j]==0&&Min>low[j])
            {
                pos=j;
                Min=low[j];
            }
        vis[pos]=1;
        for(int j=1; j<=n; j++)
            if(vis[j]==0&&low[j]>graph[pos][j]+low[pos])
            {
                low[j]=graph[pos][j]+low[pos];
            }
    }
}
int dfs(int s,int n){
    if(dp[s]>0) return dp[s];
    int ans = 0;
    for(int i=1;i<=n;i++){
        if(graph[s][i]<INF&&low[s]>low[i]&&i!=s){
            ans+=dfs(i,n);
        }
    }
    dp[s] = ans;
    return dp[s];
}
int main()
{
    while(scanf("%d",&n)!=EOF,n)
    {
        scanf("%d",&m);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++) graph[i][j]=INF;
        }
        for(int i=1; i<=m; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            graph[a][b]=graph[b][a] = c;
        }
        dijkstra(n,2);
        memset(dp,0,sizeof(dp));
        dp[2]=1;
        printf("%d
",dfs(1,n));
    }
}