A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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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没有过英语四级的渣渣 果然不能做这道题。。

理解为求最短路径的个数了  。主要就是看这句话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.   大概意思就是 如果A-B有路  并且我想从A走到B那么有条件就是A到2的距离要大于B到2的距离

思路：

1.digkstra算法求出各点到2-1的最短距离 同时存贮其它点到2的最短距离

2.记忆化搜索 从1开始 找到符合条件的路径数

#include <stdio.h>
#include <vector>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
int min_path;
int min_count;
//存贮当前点到2的最短路径
int dis[1005];
//记忆化搜索时存贮路径数
int dp[1005];
int n;
//digkstra算法和记忆化搜索时 判断是否已经走过某点
bool vis[1005];
struct node
{
	int pos;
	int cost;
	bool friend operator<(node x,node y)
	{
		return x.cost>y.cost;
	}
};
//存贮边
vector<node>map[1005];
int digkstra(int x)
{
	priority_queue<node>s;
	memset(vis,false,sizeof(vis));
	memset(dis,100,sizeof(dis));
	node temp,temp1;
	temp.pos=x;temp.cost=0;
	s.push(temp);
	while(!s.empty())
	{
		temp=temp1=s.top();s.pop();
		vis[temp.pos]=true;
		dis[temp.pos]=min(dis[temp.pos],temp.cost);
		if(temp.pos==1)
		return temp.cost;
		for(int i=0;i<map[temp.pos].size();i++)
		{
			int x=map[temp.pos][i].pos;
			int y=map[temp.pos][i].cost;
			if(!vis[x])
			{
				temp.pos=x;
				temp.cost+=y;
				s.push(temp);
			}
			temp=temp1;
		}
	}
	return -1;
}
//记忆化搜索
int dfs(int pos,int sum)
{
	if(dp[pos]) return dp[pos];
	if(pos==2) return 1;
	int result=0;
	for(int i=0;i<map[pos].size();i++)
	{
		int x=map[pos][i].pos;
		int y=map[pos][i].cost;
		//pos->x有路  并且满足pos到2的最短路径大于x到2的最短路径
		if(!vis[x]&&dis[pos]>dis[x])
		{
			vis[x]=true;
			result+=dfs(x,sum+y);
			vis[x]=false;
		}
	}
	dp[pos]=result;
	return dp[pos];
}
int main()
{
	while(~scanf("%d",&n)&&n)
	{
		int m;
		scanf("%d",&m);
		memset(map,0,sizeof(map));
		for(int i=0;i<m;i++)
		{
			int a,b,x;
			scanf("%d %d %d",&a,&b,&x);
			node temp;
			temp.pos=b;temp.cost=x;
			map[a].push_back(temp);
			temp.pos=a;
			map[b].push_back(temp);
		}
		min_path=digkstra(2);
		min_count=0;
		memset(vis,false,sizeof(vis));
		memset(dp,0,sizeof(dp));
		vis[1]=true;
		printf("%d\n",dfs(1,0));
	}
	return 0;
}