HDU 1142 A Walk Through the Forest(记忆化DFS+DIJKS最短路算法)

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9641    Accepted Submission(s): 3545

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

 

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#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define INF 100000000
#define LIM 1010
 int mp[LIM][LIM];
int n,m;
/*
题目在题意上就设置了一些小陷阱：
计数的对象不是最短路，条件是可以从A到B如果B的路径比A短（到终点），
这才是限制条件，如果我们知道每个点到终点的最短路，
那么可以想到用搜索加记忆化来实现计数问题，搜索树的构建
基于当前点与扩展子节点的关系，即d[p]>d[i]这样。

对于每个点到2的最短路，可以用D算法，也可以用BFS优先队列。。
结果存储在d[]数组中


*/
int d[LIM],vis[LIM],lowcast[LIM];
void Dijik()
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)  lowcast[i]=mp[i][2];
    d[2]=0;vis[2]=1;
    int minn,k;

    for(int i=1;i<n;i++)
    {
        minn=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&minn>lowcast[j])
            {
                minn=lowcast[j];
                k=j;
            }
        }

        vis[k]=1;
        d[k]=minn;

        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&lowcast[j]>mp[j][k]+d[k])
                lowcast[j]=mp[j][k]+d[k];
        }
    }
}
/*
void Dijik()
{
    for(int i=0;i<LIM;i++) vis[i]=MAX;
     queue<Node> seq;
     seq.push(Node(0,1));
     vis[1]=0;
     while(!seq.empty())
     {
        Node t=seq.front();
        seq.pop();
        if(t.id==2) continue;
        for(int i=1;i<=n;i++)
        {
            //if(mp[i][t.id]&&i==2) seq.push(Node(tmp,i))
            if(mp[i][t.id]==0||vis[t.id]+mp[t.id][i]>vis[i]) continue;
            int tmp=t.dis+mp[i][t.id];
            seq.push(Node(tmp,i));
            vis[i]=tmp;
           // if(i!=2)   vis[i]=1;
            newmp[t.id][i]=1;
           //cout<<t.id<<" "<<i<<endl;
        }
     }
}
*/
/*
int BFS()
{
    queue<int> seq;
    seq.push(1);
    int cnt=0;
    while(!seq.empty())
    {
        int t=seq.front();
        seq.pop();
        if(t==2) { cnt++;continue;}
        for(int i=1;i<=n;i++)
        {
            if(newmp[t][i]) seq.push(i);
        }
    }
    return cnt;
}
*/
int mem[LIM];
int mDFS(int p)
{
    if(p==2) return 1;
    if(mem[p]>0) return mem[p];
   // int &ans=mem[p];
   int ans=0;
    for(int i=1;i<=n;i++)
    {
        if(mp[i][p]<INF&&d[i]<d[p])
        {
            if(mem[i]>0) ans+=mem[i];
            else  ans+=mDFS(i);
        }
    }
    ans+=mem[p];
    mem[p]=ans;
    //mem[p]=ans;
    return ans;
}

int main()
{
    while((cin>>n)&&n)
    {
        memset(mem,0,sizeof(mem));
        cin>>m;int x,y,z;

        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
        {
            if(i==j) mp[i][j]=0;
            else mp[i][j]=INF;
        }

        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            mp[x][y]=mp[y][x]=z;
        }

        Dijik();
        printf("%d\n",mDFS(1));
    }
    return 0;
}