A1003 Emergency (25分)

一、技术总结

  1. 这是一道考查Djiktra算法的题目,主要是求解最短路径的前提下,求每个城市点权的最大值,同时最短路径的条数。
  2. 只需要在Djikstra算法的基本架构下,添加一些条件即可,点权是添加一个数组weight存储每个点的权值,然后再创建数组w,用于存储。初始化是,除了起点s,w[s]=weight[s],其余都初始化为0,使用memset(w, 0, sizeof(w)),进行初始化。
  3. 而对于最短路径的条数,设置数组num,然后初始化,num[s] = 1,其余都是初始化为0,memset(num, 0, sizeof(num))。
  4. 还有就是书写的问题,注意关系。具体参考代码。

二、参考代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 510;
const int INF = 100000000;
int n, G[maxn][maxn];
bool vis[maxn] = {false};
int d[maxn];//用于记录最短路径
int num[maxn];//用于记录最短路径的条数
int weight[maxn];//记录城市中可以调动的人数
int w[maxn];
void Djikstra(int s){
    fill(d, d+maxn, INF);
    memset(num, 0, sizeof(num));
    memset(w, 0, sizeof(w));
    //fill(num, num+maxn, 0);
    //fill(w, w+maxn, 0);
    d[s] = 0;
    num[s] = 1;
    w[s] = weight[s];
    for(int i = 0; i < n; i++){
        int u = -1, MIN = INF;
        for(int j = 0; j < n; j++){
            if(vis[j] == false && d[j] < MIN){
                u = j;
                MIN = d[j];
            }
        }
        if(u == -1) return;
        vis[u] = true;
        for(int v = 0; v < n; v++){
            if(vis[v] == false && G[u][v] != INF){
                if(d[u] + G[u][v] < d[v]){
                    d[v] = d[u] + G[u][v];
                    w[v] = w[u] + weight[v];
                    num[v] = num[u];
                }else if(d[u] + G[u][v] == d[v]){
                    num[v] += num[u];
                    if(w[u] + weight[v] > w[v]){
                        w[v] = w[u] + weight[v];
                    }
                }
            }
        }
    }
}
int main(){
    int m, c1, c2;
    scanf("%d%d%d%d", &n, &m, &c1, &c2);
    for(int i = 0; i < n; i++){
        scanf("%d", &weight[i]);
    }
    int id1, id2, L;//城市编号和他们之间的距离
    fill(G[0], G[0]+maxn*maxn, INF);
    for(int i = 0; i < m; i++){
        scanf("%d %d %d", &id1, &id2, &L);
        G[id1][id2] = L;
        G[id2][id1] = L;
    }
    Djikstra(c1);
    printf("%d %d", num[c2], w[c2]);
    return 0;
}

原文地址:https://www.cnblogs.com/tsruixi/p/12384951.html

时间: 2024-07-31 18:54:52

A1003 Emergency (25分)的相关文章

【PAT甲级】1003 Emergency (25分)

1003 Emergency (25分) As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between an

A1003. Emergency (25)

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are

PAT 1003 Emergency (25)(25 分)

1003 Emergency (25)(25 分) As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road betwe

1003 Emergency(25 分)C语言版本(提问求解答)

1003 Emergency(25 分) As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between an

4-9 二叉树的遍历 (25分)

4-9 二叉树的遍历   (25分) 输出样例(对于图中给出的树): Inorder: D B E F A G H C I Preorder: A B D F E C G H I Postorder: D E F B H G I C A Levelorder: A B C D F G I E H 代码:(都是遍历的算法) 1 // 4-9 二叉树的遍历 2 // 3 // Created by Haoyu Guo on 04/02/2017. 4 // Copyright ? 2017 Haoy

5-24 树种统计 (25分)

5-24 树种统计   (25分) 随着卫星成像技术的应用,自然资源研究机构可以识别每一棵树的种类.请编写程序帮助研究人员统计每种树的数量,计算每种树占总数的百分比. 输入格式: 输入首先给出正整数N(\le 10^5≤10?5??),随后N行,每行给出卫星观测到的一棵树的种类名称.种类名称由不超过30个英文字母和空格组成(大小写不区分). 输出格式: 按字典序递增输出各种树的种类名称及其所占总数的百分比,其间以空格分隔,保留小数点后4位. 输入样例: 29 Red Alder Ash Aspe

5-20 表达式转换 (25分)

5-20 表达式转换 (25分) 算术表达式有前缀表示法.中缀表示法和后缀表示法等形式.日常使用的算术表达式是采用中缀表示法,即二元运算符位于两个运算数中间.请设计程序将中缀表达式转换为后缀表达式. 输入格式: 输入在一行中给出不含空格的中缀表达式,可包含+.-.*.\以及左右括号( ),表达式不超过20个字符. 输出格式: 在一行中输出转换后的后缀表达式,要求不同对象(运算数.运算符号)之间以空格分隔,但结尾不得有多余空格. 输入样例: 2+3*(7-4)+8/4 输出样例: 2 3 7 4

5-17 汉诺塔的非递归实现 (25分)

5-17 汉诺塔的非递归实现   (25分) 借助堆栈以非递归(循环)方式求解汉诺塔的问题(n, a, b, c),即将N个盘子从起始柱(标记为"a")通过借助柱(标记为"b")移动到目标柱(标记为"c"),并保证每个移动符合汉诺塔问题的要求. 输入格式: 输入为一个正整数N,即起始柱上的盘数. 输出格式: 每个操作(移动)占一行,按柱1 -> 柱2的格式输出. 输入样例: 3 输出样例: a -> c a -> b c -&g

5-3 树的同构 (25分)

5-3 树的同构   (25分) 给定两棵树T1和T2.如果T1可以通过若干次左右孩子互换就变成T2,则我们称两棵树是"同构"的.例如图1给出的两棵树就是同构的,因为我们把其中一棵树的结点A.B.G的左右孩子互换后,就得到另外一棵树.而图2就不是同构的. 图1 图2 现给定两棵树,请你判断它们是否是同构的. 输入格式: 输入给出2棵二叉树树的信息.对于每棵树,首先在一行中给出一个非负整数NN (\le 10≤10),即该树的结点数(此时假设结点从0到N-1N?1编号):随后NN行,第i