HDU 1142 A Walk Through the Forest(dijkstra+记忆化DFS)

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9945    Accepted Submission(s): 3623

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

Sample Output

2
4

Source

University of Waterloo Local Contest 2005.09.24

分析：

题目都没有看懂。。。

英语渣

题目意思：

给你一个图，找最短路，但是有一个条件，如果a，b之间右路，且你要选这条路，必须保证a到终点的所有路都小于b到终点的所有路，问这样的路的条数

思路：

1为起点，2为终点，必须满足那个条件，先求每个点到终点的最短路距离，保存在dis数组

然后从起点开始DFS，看看满足条件的路的条数

注意要采用记忆化搜索（备忘录）

dp【1】就是从起点到终点满足条件的路径数目

dfs自己还不会写，看的别人的。。。。

mmp

#include<bits/stdc++.h>
using namespace std;
#define max_v 1005
#define INF 99999
int e[max_v][max_v];
int n,m;
int used[max_v];
int dis[max_v];
int dp[max_v];
void init()
{
    memset(used,0,sizeof(used));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            e[i][j]=INF;
        }
        dis[i]=INF;
    }
}
void Dijkstra(int s)
{
    for(int i=1;i<=n;i++)
    {
        dis[i]=e[s][i];
        dp[i]=-1;
    }
    dis[s]=0;
    for(int i=1;i<=n;i++)
    {
        int index,mindis=INF;
        for(int j=1;j<=n;j++)
        {
            if(used[j]==0&&dis[j]<mindis)
            {
                mindis=dis[j];
                index=j;
            }
        }
        used[index]=1;
        for(int j=1;j<=n;j++)
        {
            if(dis[index]+e[index][j]<dis[j])
                dis[j]=dis[index]+e[index][j];
        }
    }
}
int dfs(int x)
{
    int sum=0;
    if(dp[x]!=-1)
        return dp[x];
    if(x==2)
        return 1;
    for(int i=1;i<=n;i++)
    {
        if(e[x][i]!=INF&&dis[x]>dis[i]&&i!=x)
        {
            sum+=dfs(i);
        }
    }
    dp[x]=sum;
    return dp[x];
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        scanf("%d",&m);
        init();
        for(int i=0;i<m;i++)
        {
            int a,b,c;
            scanf("%d %d %d",&a,&b,&c);
            e[a][b]=e[b][a]=c;
        }
        Dijkstra(2);
        printf("%d\n",dfs(1));
    }
}