HDU 1142 A Walk Through the Forest （记忆化搜索+Dijkstra算法）

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7583    Accepted Submission(s): 2791

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

 

Source

University of Waterloo Local Contest 2005.09.24

 

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Eddy

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1142

题意：

输入n,m,分别表示顶点数和路径数，接下来m行有三个数，代表起点，终点，路径长度。

问：从‘1’开始走，到达‘2’有多少种情况。

路径要求：如果要从A点到达B，必须满足B点到终点的最短路径长度必须大于A到终点的最短路径长度。

做法：以‘2’为起点，用Dijkstra算法求出到其他点的最短路径长度，接下来就是记忆化搜索了。

AC代码：

#include <iostream>#include <cstring>using namespace std;const int maxn=1005;int dis[maxn];int a[maxn][maxn];bool vis[maxn];int dp[maxn];const int INF=0x3f3f3f3f;int n,m;void DIJ(int x){    for(int i=1; i<=n; i++)    {        dis[i]=a[x][i];    }    memset(vis,false,sizeof(vis));    vis[x]=true;    dis[x]=0;    int p;    for(int i=1; i<n; i++)    {        int minn=INF;        for(int j=1; j<=n; j++)        {            if(!vis[j]&&dis[j]<minn)                minn=dis[p=j];        }        if(minn==INF) break;        vis[p]=true;        for(int j=1; j<=n; j++)        {            if(!vis[j]&&dis[p]+a[p][j]<dis[j])                dis[j]=dis[p]+a[p][j];        }    }}int DFS(int x){    int sum=0;    if(dp[x]!=-1) return dp[x];    for(int i=1; i<=n; i++)    {        if(a[x][i]!=INF&&dis[x]>dis[i])            sum+=DFS(i);    }    return dp[x]=sum;}int main(){    while(cin>>n,n)    {        cin>>m;        int x,y,z;        memset(a,INF,sizeof(a));        while(m--)        {            cin>>x>>y>>z;            a[x][y]=a[y][x]=z;        }        DIJ(2);        memset(dp,-1,sizeof(dp));        dp[2]=1;        cout<<DFS(1)<<endl;    }    return 0;}