zoukankan      html  css  js  c++  java
  • PAT 甲级 1030 Travel Plan

    https://pintia.cn/problem-sets/994805342720868352/problems/994805464397627392

    A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

    Input Specification:

    Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤) is the number of cities (and hence the cities are numbered from 0 to N1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

    City1 City2 Distance Cost
    

    where the numbers are all integers no more than 500, and are separated by a space.

    Output Specification:

    For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

    Sample Input:

    4 5 0 3
    0 1 1 20
    1 3 2 30
    0 3 4 10
    0 2 2 20
    2 3 1 20
    

    Sample Output:

    0 2 3 3 4

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    #define inf 0x3f3f3f3f
    
    int N, M, S, D;
    int mp[550][550], cost[550][550];
    int dis1[550],dis2[550],dis3[550];
    int vis[550];
    int pre[550];
    
    int MinStep, out = INT_MAX;
    
    void dijkstra(int S, int dis[]) {
        dis[S] = 0;
        memset(vis, 0, sizeof(vis));
        int temp = S;
    
        for(int i = 0; i < N; i ++) {
            int minn = inf;
            for(int j = 0; j < N; j ++) {
                if(dis[j] < minn && vis[j] == 0) {
                    minn = dis[j];
                    temp = j;
                }
            }
            vis[temp] = 1;
            for(int k = 0; k < N; k ++)
                if(vis[k] == 0 && mp[temp][k] != inf) {
                    if(dis[k] > mp[temp][k] + dis[temp])
                        dis[k] = mp[temp][k] + dis[temp];
            }
        }
    }
    
    void dijkstra(int S) {
    
        dis3[S] = 0;
        memset(vis, 0, sizeof(vis));
        int temp = S;
    
        for(int i = 0; i < N; i ++) {
            int minn = inf;
            for(int j = 0; j < N; j ++) {
                if(dis3[j] < minn && vis[j] == 0) {
                    minn = dis3[j];
                    temp = j;
                }
            }
            vis[temp] = 1;
            for(int k = 0; k < N; k ++)
                if(mp[temp][k] + dis1[temp] + dis2[k] == MinStep)
                    if(vis[k] == 0 && cost[temp][k] != inf)
                        if(dis3[k] > cost[temp][k] + dis3[temp]){
                            dis3[k] = cost[temp][k] + dis3[temp];
                            pre[k] = temp;
                        }
        }
    }
    
    
    void output(int d){
        if(d==-1) return ;
        output(pre[d]);
        printf("%d ", d);
    }
    
    int main() {
    
        memset(pre,-1, sizeof(pre));
        memset(vis, 0, sizeof(vis));
        memset(cost,inf, sizeof(cost));
    
        memset(dis1, inf, sizeof(dis1));
        memset(dis2, inf, sizeof(dis2));
        memset(dis3, inf, sizeof(dis3));
    
        memset(mp, inf, sizeof(mp));
    
        scanf("%d%d%d%d", &N, &M, &S, &D);
        for(int i = 0; i < M; i ++) {
            int st, en, dist, val;
            scanf("%d%d%d%d", &st, &en, &dist, &val);
            mp[st][en] = mp[en][st] = min(dist, mp[en][st]);
            cost[st][en] = cost[en][st] = min(val, cost[st][en]);
        }
    
        dijkstra(S, dis1);
        dijkstra(D, dis2);
    
        MinStep = dis1[D];
    
        dijkstra(S);
    
        output(D);
        printf("%d %d
    ", dis1[D], dis3[D]);
        return 0;
    }
    

      两遍 dijkstra 一上午经历了无数遍点开题目又退出 枯了 睡一会清醒清醒再来写吧

  • 相关阅读:
    PHP 小tip .(@)符号和 php if 赋值
    PHP定义字符串时单引号和双引号的区别
    笔记本 windows 10 安装
    unix_12c_db_init
    教你如何使用php session
    js和 php 介绍
    PHP 简单答题系统
    PHP list() 函数
    PHP集成支付宝快速实现充值功能
    eclipse如何导入PHP的项目
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10369432.html
Copyright © 2011-2022 走看看