ZOJ2182&&POJ1966-Cable TV Network(求顶点连通度)
来源:互联网 发布:java poi 导出xlsx 编辑:程序博客网 时间:2024/05/16 10:14
The interconnection of the relays in a cable TV network is bi-directional. The network is connected if there is at least one interconnection path between each pair of relays present in the network. Otherwise the network is disconnected. An empty network or a network with a single relay is considered connected. The safety factor f of a network with n relays is:
1. n, if the net remains connected regardless the number of relays removed from the net.
2. The minimal number of relays that disconnect the network when removed.
For example, consider the nets from figure 1, where the circles mark the relays and the solid lines correspond to interconnection cables. The network (a) is connected regardless the number of relays that are removed and, according to rule (1), f=n=3. The network (b) is disconnected when 0 relays are removed, hence f=0 by rule (2). The network (c) is disconnected when the relays 1 and 2 or 1 and 3 are removed. The safety factor is 2.
Input
Write a program that reads several data sets from the standard input and computes the safety factor for the cable networks encoded by the data sets. Each data set starts with two integers: 0<=n<=50,the number of relays in the net, and m, the number of cables in the net. Follow m data pairs (u,v), u < v, where u and v are relay identifiers (integers in the range 0..n-1). The pair (u,v) designates the cable that interconnects the relays u and v. The pairs may occur in any order.Except the (u,v) pairs, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.
Output
For each data set, the program prints on the standard output, from the beginning of a line, the safety factor of the encoded net.
Sample Input
0 0
1 0
3 3 (0,1) (0,2) (1,2)
2 0
5 7 (0,1) (0,2) (1,3) (1,2) (1,4) (2,3) (3,4)
Sample Output
0
1
3
0
2
Source: Southeastern Europe 2004
题意:给出n个点和m条边,求顶点连通度
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;int mp[105][2];int flag[105][105];struct Edge{ int u,v,c,next;}edge[10000];int n,m,cnt;int dist[3000];int s[3000];int que[3000],sta[3000];void addedge(int u,int v,int c){ edge[cnt].u=u; edge[cnt].v=v; edge[cnt].c=c; edge[cnt].next=s[u]; s[u]=cnt++; edge[cnt].u=v; edge[cnt].v=u; edge[cnt].c=0; edge[cnt].next=s[v]; s[v]=cnt++;}int dinic(int ss,int tt){ int ans=0; while(true) { int left,right,u,v; memset(dist,-1,sizeof(dist)); left=right=0; que[right++]=ss; dist[ss]=0; while(left<right) { u=que[left++]; for(int i=s[u];~i;i=edge[i].next) { u=edge[i].u; v=edge[i].v; if(edge[i].c>0&&dist[v]==-1) { dist[v]=dist[u]+1; que[right++]=v; if(v==tt) { left=right; break; } } } } if(dist[tt]==-1) break; int top=0,now=ss; while(true) { if(now!=tt) { int i; for(i=s[now];~i;i=edge[i].next) if(edge[i].c>0&&dist[edge[i].v]==dist[edge[i].u]+1) break; if(i!=-1) { sta[top++]=i; now=edge[i].v; } else { if(top==0) break; dist[edge[sta[--top]].v]=-1; now=edge[sta[top]].u; } } else { int flow=INF,k; for(int i=0; i<top; i++) { if(flow>edge[sta[i]].c) { flow=edge[sta[i]].c; k=i; } } ans+=flow; for(int i=0; i<top; i++) { edge[sta[i]].c-=flow; edge[sta[i]^1].c+=flow; } now=edge[sta[k]].u; top=k; } } } return ans;}void build(int x,int y,int ver,int n,int m){ cnt=0; memset(s,-1,sizeof s); for(int i=1; i<=n; i++) addedge(i,i+n,1); for(int i=0; i<m; i++) { addedge(mp[i][0]+n,mp[i][1],INF); addedge(mp[i][1]+n,mp[i][0],INF); } addedge(x,x+n,INF); addedge(y,y+n,INF);}int main(){ while(~scanf("%d%d",&n,&m)) { int a,b; int ver=n*2+1; memset(mp,0,sizeof mp); memset(flag,0,sizeof flag); for(int i=0; i<m; i++) { scanf(" (%d,%d)",&a,&b); a++,b++; mp[i][0]=a,mp[i][1]=b; flag[a][b]=1; } if(m==0) { if(n==1) printf("1\n"); else printf("0\n"); continue; } int ans=INF,flag1=0; for(int i=1; i<=n; i++) { for(int j=i+1; j<=n; j++) { if(!flag[i][j]) { flag1=1; build(i,j,ver,n,m); ans=min(ans,dinic(i,j+n)); } } } if(flag1) printf("%d\n",ans); else printf("%d\n",n); } return 0;}
- ZOJ2182&&POJ1966-Cable TV Network(求顶点连通度)
- POJ1966 Cable TV Network 点连通度
- Cable TV Network-POJ1966图的连通度
- POJ 1966 Cable TV Network(顶点连通度)
- 【poj1966】Cable TV Network 无向图点连通度(最大流)
- poj1966 - Cable TV Network
- POJ1966--Cable TV Network
- POJ1966 Cable TV Network
- poj1966求顶点连通度
- poj 1966 zoj 2182 Cable TV Network(无向图顶点连通度(sap求最大流))
- POJ 1966 Cable TV Network 顶点连通度的求解
- POJ 1966 / ZOJ 2182 : Cable TV Network - 顶点连通度
- poj 1966 Cable TV Network 顶点连通度-最大流
- POJ1966.Cable TV Network——无向图的点连通度
- POJ 1966 Cable TV Network(无向图的顶点连通度)
- poj1966 Cable TV Network 最大流
- POJ--1966--Cable TV Network【无向图顶点连通度】
- 【连通图|点连通度】POJ-1966 Cable TV Network
- Linux C 连接mysql
- 1008. 数组元素循环右移问题 (20)
- CentOS7编译boost1.63.0
- POI操作Excel:cell的背景颜色类型
- 前端程序员需要知道的7种新型的CSS长度单位
- ZOJ2182&&POJ1966-Cable TV Network(求顶点连通度)
- 前端设计简约
- 实例介绍机械臂运动规划及前沿研究方向(附PPT+视频)|硬创公开课
- 文章目录
- 猜数字小游戏
- nyistOJ-“炫舞家“ST(DP)
- CSDN日报20170325——《一篇文章教会你,如何做到招聘要求中的“要有扎实的Java基础”。》
- Android性能优化系列之Bitmap图片优化
- 设计模式之工厂方法模式