pat 1013. Battle Over Cities (25)
来源:互联网 发布:淘宝短链接生成 编辑:程序博客网 时间:2024/06/04 17:51
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
这题我用的是并查集,每次都构建一个并查集,父节点是自己的节点个数便是连通分量的个数cnt。去掉的节点是个连通分量,其余的有cnt-1个连通分量需要cnt-2条路。当然 cnt可能是1或0;这题M很大,我开十万的爆了,开百万才行。
#include<cstdio>#include<cstring>using namespace std;int p[10000];int findp(int x){ if (x!=p[x]) p[x]=findp(p[x]); return p[x];}void merge(int x,int y){ int px=findp(x); int py=findp(y); if (px==py) return; p[px]=py;}int main(){ int n,i,j,k,m; int x[1000000],y[1000000]; scanf("%d%d%d",&n,&m,&k); for (i=0;i<m;i++) scanf("%d%d",&x[i],&y[i]); for (i=0;i<k;i++) { int now; scanf("%d",&now); for (j=1;j<=n;j++) p[j]=j; for (j=0;j<m;j++) { if (x[j]!=now&&y[j]!=now) merge(x[j],y[j]); } int cnt=0; for (j=1;j<=n;j++) { if (p[j]==j) cnt++; } cnt-=2; if (cnt<0) cnt=0; printf("%d\n",cnt); } return 0;}
- PAT 1013. Battle Over Cities (25) DFS
- 1013. Battle Over Cities (25)-PAT
- pat 1013. Battle Over Cities (25)
- 【PAT】1013. Battle Over Cities (25)
- PAT: 1013. Battle Over Cities (25)
- PAT A 1013. Battle Over Cities (25)
- pat 1013. Battle Over Cities (25)
- PAT 1013. Battle Over Cities (25)
- PAT 1013. Battle Over Cities (25)
- PAT 1013. Battle Over Cities (25)
- PAT-A 1013. Battle Over Cities (25)
- PAT 1013. Battle Over Cities (25)
- PAT 1013. Battle Over Cities (25)
- PAT 1013. Battle Over Cities (25)
- 【PAT甲级】1013. Battle Over Cities (25)
- PAT A 1013. Battle Over Cities (25)
- PAT(A) - 1013. Battle Over Cities (25)
- 1013. Battle Over Cities (25) PAT甲级
- Retrofit 原理简析
- OpenCV 3.2 Tracking 物体跟踪
- JavaWeb Filter 过滤器的机制 与使用 请求重定向与请求转发的区别 跳转过滤,过滤器dispatcher REQUEST与FORWARD区别
- Java_基础—异常的注意事项及如何使用异常处理
- ArcGIS Engine 节点编辑,实现要素拖动、编辑、节点删除
- pat 1013. Battle Over Cities (25)
- C# yyyyMMddHHmmss格式转换DateTime
- CSS鼠标滑动显示标题全部移开显示几个字符
- leetcode(4) Median of Two Sorted Arrays
- faster-rcnn 之 bbox_transform_inv(boxes, deltas)
- ThreadPool.QueueUserWorkItem的性能问题
- 产品经理与项目经理的区别
- Ubuntu 终端常用命令
- 仿京东底部导航栏实现