hdu 1142 A Walk Through the Forest (digkstra+記憶化搜索)

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7322Accepted Submission(s): 2685


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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沒有過英語四級的渣渣 果然不能做這道題。。

理解為求最短路徑的個數了 。主要就是看這句話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 並且我想從A走到B那麼有條件就是A到2的距離要大於B到2的距離

思路：

1.digkstra算法求出各點到2-1的最短距離 同時存貯其它點到2的最短距離

2.記憶化搜索 從1開始 找到符合條件的路徑數

 

#include <stdio.h>
#include <vector>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
int min_path;
int min_count;
//存貯當前點到2的最短路徑 
int dis[1005];
//記憶化搜索時存貯路徑數 
int dp[1005];
int n;
//digkstra算法和記憶化搜索時 判斷是否已經走過某點 
bool vis[1005];
struct node
{
	int pos;
	int cost;
	bool friend operator<(node x,node y)
	{
		return x.cost>y.cost;
	}
};
//存貯邊 
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++) i="0;i<map[pos].size();i++)" int="" pos="=2)" pos-="" result="0;" return="" temp="temp1;" temp.pos="x;" x="map[temp.pos][i].pos;" y="map[temp.pos][i].cost;">x有路  並且滿足pos到2的最短路徑大於x到2的最短路徑 
		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++) d="" int="" min_count="0;" min_path="digkstra(2);" node="" pre="" return="" temp.pos="b;temp.cost=x;"><p>
</p><p>
</p>
   
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