PAT-Top1001. Battle Over Cities - Hard Version (35)
在敌人占领之前由城市和公路构成的图是连通图。在敌人占领某个城市之后所有通往这个城市的公路就会被破坏,接下来可能需要修复一些其他被毁坏的公路使得剩下的城市能够互通。修复的代价越大,意味着这个城市越重要。如果剩下的城市无法互通,则说明代价无限大,这个城市至关重要。最后输出的是代价最大的城市序号有序列表。借助并查集和Kruskal算法(最小生成树算法)来解决这个问题。
//#include "stdafx.h"
#include <iostream>
#include <algorithm>
#include <vector> using namespace std; struct edge { // edge struct
int u, v, cost;
};
vector<edge> edges; // the number of edges is greater than 500 far and away int cmp(edge a, edge b) { // sort rule
return a.cost < b.cost;
} int parent[]; // union-find set void initParent(int n) { // initialize union-find set
int i;
for(i = ; i <= n; i++) {
parent[i] = -; // a minus means it is a root node and its absolute value represents the number of the set
}
} int findRoot(int x) { // find the root of the set
int s = x;
while(parent[s] > ) {
s = parent[s];
} int temp;
while(s != x) { // compress paths for fast lookup
temp = parent[x];
parent[x] = s;
x = temp;
} return s;
} void unionSet(int r1, int r2) { // union sets. More concretely, merge a small number of set into a large collection
int sum = parent[r1] + parent[r2];
if(parent[r1] > parent[r2]) {
parent[r1] = r2;
parent[r2] = sum;
} else {
parent[r2] = r1;
parent[r1] = sum;
}
} int maxw = ; // max cost
bool infw; // infinite cost int kruskal(int n, int m, int out) { // Kruskal algorithm to get minimum spanning tree
initParent(n); int u, v, r1, r2, num = , i, w = ;
for (i = ; i < m; i++) {
u = edges[i].u;
v = edges[i].v; if (u == out || v == out) {
continue;
} r1 = findRoot(u);
r2 = findRoot(v); if (r1 != r2) {
unionSet(r1, r2);
num++; if (edges[i].cost > ) { // only consider the cost which is not zero
w += edges[i].cost;
} if (num == n - ) {
break;
}
}
} //printf("num %d\n", num);
if (num < n - ) { // spanning tree is not connected
w = -; // distinguish the situation of the occurrence of infinite cost if (!infw) { // when infinite cost first comes out
infw = true;
}
} return w;
} int main() {
int n, m;
scanf("%d%d", &n, &m); int i, status;
edge e;
for (i = ; i < m; i++) {
scanf("%d%d%d%d", &e.u, &e.v, &e.cost, &status);
if (status == ) {
e.cost = ;
} edges.push_back(e);
} if (m > ) {
sort(edges.begin(), edges.end(), cmp);
} int curw, res[], index = ;
for (i = ; i <= n; i++) { // traverse all vertices to obtain the target vertex
curw = kruskal(n, m, i);
if (!infw) { // when infinite cost doesn't come out
if (curw < maxw) {
continue;
} if (curw > maxw) {
index = ;
maxw = curw;
}
res[index++] = i;
} else { // otherwise
if (curw < ) {
if (maxw > ) {
maxw = -;
index = ;
} res[index++] = i;
}
}
} if (index > ) {
for (i = ; i < index; i++) {
if (i > ) {
printf(" ");
}
printf("%d", res[i]);
}
} else {
printf("");
}
printf("\n"); system("pause");
return ;
}
参考资料
pat-top 1001. Battle Over Cities - Hard Version (35)
PAT-Top1001. Battle Over Cities - Hard Version (35)的更多相关文章
- PAT 1013 Battle Over Cities
1013 Battle Over Cities (25 分) It is vitally important to have all the cities connected by highway ...
- PAT 1013 Battle Over Cities(并查集)
1013. Battle Over Cities (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue It ...
- pat 1013 Battle Over Cities(25 分) (并查集)
1013 Battle Over Cities(25 分) It is vitally important to have all the cities connected by highways i ...
- PAT 1013 Battle Over Cities (dfs求连通分量)
It is vitally important to have all the cities connected by highways in a war. If a city is occupied ...
- PAT A1013 Battle Over Cities (25 分)——图遍历,联通块个数
It is vitally important to have all the cities connected by highways in a war. If a city is occupied ...
- PAT 1013 Battle Over Cities DFS深搜
It is vitally important to have all the cities connected by highways in a war. If a city is occupied ...
- pat1001. Battle Over Cities - Hard Version 解题报告
/**题目:删去一个点,然后求出需要增加最小代价的边集合生成连通图思路:prim+最小堆1.之前图中未破坏的边必用,从而把两两之间可互达的点集合 合并成一个点2.求出不同点集合的最短距离,用prim+ ...
- 「日常训练」Battle Over Cities - Hard Version(PAT-TOP-1001)
题意与分析 题意真的很简单,实在不想讲了,简单说下做法吧. 枚举删除每个点,然后求最小生成树,如果这个路已经存在那么边权就是0,否则按照原来的处理,之后求花费,然后判整个图是否联通(并查集有几个roo ...
- PAT_A1013#Battle Over Cities
Source: PAT A1013 Battle Over Cities (25 分) Description: It is vitally important to have all the cit ...
随机推荐
- Leetcode刷题第004天
class Solution { public: int findKthLargest(vector<int>& nums, int k) { , nums.size()-, k) ...
- xxl系列部署启动通用办法
http://10.10.6.186:8080/xxl-job-admin # 编译mvn compile # 清理mvn clean # 打包mvn package # 先清理后编译mvn clea ...
- JavaBean toString() - 将bean对象打印成字符串
JavaBean toString方式 https://www.cnblogs.com/thiaoqueen/p/7086195.html //方法一:自动生成 @Override public St ...
- elk服务器和运维服务器的IPTABLES
--运维服务器 iptables -P INPUT ACCEPT iptables -F iptables -X iptables -Z iptables -A INPUT -i lo -j ACCE ...
- javascript 对象(四)
一.对象概述 对象中包含一系列的属性,这些属性是无序的.每个属性都有一个字符串key和对应的value. var obj={x:1,y:2}; obj.x; obj.y; 1.为什么属性的key必须是 ...
- mysql8.0.11绿色版安装教程
解压到安装目录 在根目录建立data文件夹 建立my.ini文件 代码如下 # Other default tuning values # MySQL Server Instance Configur ...
- Java第三阶段学习(七、线程池、多线程)
一.线程池 1.概念: 线程池,其实就是一个容纳多个线程的容器,其中的线程可以重复使用,省去了频繁创建线程对象的过程,无需反复创建线程而消耗过多资源,是JDK1.5以后出现的. 2.使用线程池的方式- ...
- Python学习(十一) —— 模块和包
一.模块 一个模块就是一个包含了python定义和声名的文件,文件名就是模块名加上.py后缀. import加载的模块分为四个通用类别: 1.使用python编写的代码(.py文件) 2.已被编译为共 ...
- Spring Data Redis实现消息队列——发布/订阅模式
一般来说,消息队列有两种场景,一种是发布者订阅者模式,一种是生产者消费者模式.利用redis这两种场景的消息队列都能够实现. 定义:生产者消费者模式:生产者生产消息放到队列里,多个消费者同时监听队列, ...
- ibatis的queyrForList和queryForMap区别
https://blog.csdn.net/z69183787/article/details/47360825 https://blog.csdn.net/zyq527758142/article/ ...