A Walk Through the Forest

最新推荐文章于 2025-01-28 00:20:42 发布

AC1048900581

最新推荐文章于 2025-01-28 00:20:42 发布

阅读量477

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/Burning_newbie/article/details/48783681

ACM 专栏收录该内容

95 篇文章

订阅专栏

本文介绍了一个经典的图论问题——寻找从办公室到家的不同路径数量。通过使用记忆化搜索改进算法效率，解决了Jimmy每天想要走不同路线回家的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

一开始广搜超时了，后来改成记忆化搜索

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6889 Accepted Submission(s): 2534

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

Sample Output

2
4

#include <cstdio>
#include <vector>
#include <cstring>
#include <queue>
using namespace std;
struct node{
    int v,w;
    node(int vv,int ww):v(vv),w(ww){}
    node(){}
    friend bool operator <(const node &a,const node &b){
        return a.w>b.w;
    }
};
vector<vector<node> >g;
int dist[1010];
bool used[1010];
int s[200010];
int di[1010];
void dis(int s){
    priority_queue<node> q;
    q.push(node(s,0));
    while(!q.empty()){
        node p=q.top();
        q.pop();
        int u=p.v;
       // printf("%d\n",u);
        if(used[u])continue;
        used[u]=true;
        dist[u]=p.w;
        for(int i=0;i<g[u].size();i++){
            int v=g[u][i].v;
            int w=g[u][i].w;
            if(!used[v]){
                q.push(node(v,p.w+w));
            }
        }
    }
}
int dfs(int u){
    if(u==2)return 1;
    if(di[u])return di[u];
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i].v;
        if(dist[v]<dist[u])
            di[u]+=dfs(v);
    }
    return di[u];
}
int main()
{
    int n,m;
    while(scanf("%d",&n)!=EOF){
        if(n==0)break;
        memset(dist,0,sizeof(dist));
        memset(used,0,sizeof(used));
        memset(di,0,sizeof(di));
        scanf("%d",&m);
        g.clear();
        g.resize(n+10);
        for(int i=0;i<m;i++){
            int a,b,d;
            scanf("%d%d%d",&a,&b,&d);
            g[a].push_back(node(b,d));
            g[b].push_back(node(a,d));
        }
        dis(2);
        //printf("hello");
        printf("%d\n",dfs(1));
    }
}