A Walk Through the Forest

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

 

 

題意：

輸入n,m,分別表示頂點數和路徑數，接下來m行有三個數，代表起點，終點，路徑長度。   問：從‘1’開始走，到達‘2’有多少種情況。   路徑要求：如果要從A點到達B，必須滿足B點到終點的最短路徑長度必須大於A到終點的最短路徑長度。   做法：以‘2’為起點，用Dijkstra算法求出到其他點的最短路徑長度，接下來就是記憶化搜索了。  
 

AC代碼：

#include 
#include 
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]=false;
    dis[x]=0;
    int p;
    for(int i=1; idis[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<