1013. Battle Over Cities (25)
来源:互联网 发布:电脑面部识别软件 编辑:程序博客网 时间:2024/05/16 05:15
It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.
For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. Then if city1 is occupied by the enemy, we must have 1 highway repaired, that is the highway city2-city3.
Input
Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.
Output
For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.
Sample Input
3 2 3
1 2
1 3
1 2 3
Sample Output
1
0
0
找到有多少个连通分量N,需要添加的就是N-1条,使用DFS。
#include <cstdio>#include <iostream>using namespace std;int N;int v[1001][1001];bool visit[1001];void dfs(int node){ visit[node] = true; for (int i = 1; i <= N; i++) { if (visit[i] == false && v[node][i] == 1) dfs(i); }}int main(){ int M, K, a, b; scanf("%d %d %d",&N,&M,&K); for (int i = 0; i < M; i++) { scanf("%d %d", &a, &b); v[a][b] = v[b][a] = 1; } for (int i = 0; i < K; i++) { fill(visit, visit + 1001, false); scanf("%d", &a); int count = 0; visit[a] = true; for (int j = 1; j <= N; j++) { if (visit[j] == false) { dfs(j); count++; } } printf("%d\n", count - 1); } return 0;}
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25)
- Javascript对象
- 强化学习读书笔记
- Java 重写(Override)与重载(Overload)
- jquery.validate使用攻略
- 详解正则表达式
- 1013. Battle Over Cities (25)
- 【一图流】_01_一张图看懂 Android 系统的开机流程:
- 从jvm的角度来看单例模式
- gcc -lm 选项
- eclipse如何设置自动去除多余引入的类?
- 四种类型转换
- canvas的宽度设置
- 树的层次遍历【队列的应用】
- 《Java高并发程序设计》总结--5. 并行模式与算法