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HDU 1142 A Walk Through the Forest (Dijkstra + 记忆化搜索 好题)

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A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6350    Accepted Submission(s): 2332



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
 

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1142

题目大意:一个人办公室在点1,家在点2,他要从办公室回家,他从点A到点B当且仅当从B到家的距离必任意一点从A到家的都小,求他回家路线的方案数

题目分析:好题,先用最短路预处理出到家的最短路,这里用Dijkstra算法,然后就是从办公室开始记忆化搜索



#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const INF = 0x3fffffff;
int const MAX = 1005;
int dis[MAX], dp[MAX];
int map[MAX][MAX];
bool vis[MAX];
int n, m;

void Dijkstra(int v0)
{
    memset(vis, false, sizeof(vis));
    for(int i = 1; i <= n; i++)
        dis[i] = INF;
    for(int i = 1; i <= n; i++)
        dis[i] = map[v0][i];
    vis[v0] = true;
    dis[v0] = 0;
    for(int i = 1; i < n; i++)
    {
        int u = 0, mi = INF;
        for(int j = 1; j <= n; j++)
        {
            if(!vis[j] && dis[j] < mi)
            {
                mi = dis[j];
                u = j;
            }
        }
        vis[u] = true;
        for(int k = 1; k <= n; k++)
            if(!vis[k])
                dis[k] = min(dis[k], dis[u] + map[u][k]);
    }
}

int DFS(int x)
{
    if(dp[x])
        return dp[x];
    if(x == 2)
        return 1;
    int tmp = 0;
    for(int i = 1; i <= n; i++)
        if(dis[x] > dis[i] && map[x][i] != INF)
            tmp += DFS(i);
    return dp[x] = tmp;
}

int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        scanf("%d", &m);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                map[i][j] = INF;
        for(int i = 0; i < m; i++)
        {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            map[u][v] = map[v][u] = w;
        }
        Dijkstra(2);
        memset(dp, 0, sizeof(dp));
        printf("%d\n", DFS(1));
    }
}


HDU 1142 A Walk Through the Forest (Dijkstra + 记忆化搜索 好题)

原文:http://blog.csdn.net/tc_to_top/article/details/45276461

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