nocow上的题解很好。 http://www.nocow.cn/index.php/USACO/schlnet

如何求强连通分量呢?对于此题,可以直接先用floyd,然后再判断。

----------------------------------------------------------------------------------

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define rep(i,r) for(int i=0;i<r;i++)
#define clr(x,c) memset(x,c,sizeof(x))
#define Rep(i,l,r) for(int i=l;i<r;i++)
using namespace std;
const int maxn=100+5;
int p[maxn];
int map[maxn][maxn];
int in[maxn],out[maxn];
bool ok[maxn];
int n;
void init() {
clr(map,0); clr(in,0); clr(out,0); clr(ok,0);
cin>>n;
rep(i,n) p[i]=i;
int t;
rep(i,n)
while(scanf("%d",&t) && t) map[i][--t]=1;
}
int find(int x) { return x==p[x] ? x:p[x]=find(p[x]); }
void work() {
rep(k,n)
   rep(i,n)
       rep(j,n) if(map[i][k] && map[k][j]) map[i][j]=1;
       
rep(i,n)
   Rep(j,i+1,n) if(map[i][j] && map[j][i]) p[i]=find(j);
rep(i,n) {
int x=find(i);
ok[x]=1;
rep(j,n) {
int y=find(j);
if(x==y) continue;
if(map[i][j]) out[x]++;
if(map[j][i]) in[x]++;
}
}
int cnt[2]={0,0},pd=-1;
rep(i,n) if(ok[i]) {
pd++;
if(!in[i]) cnt[0]++;
if(!out[i]) cnt[1]++;
}
if(pd) printf("%d\n%d\n",cnt[0],max(cnt[0],cnt[1]));
else printf("1\n0\n");
}
int main()
{
freopen("schlnet.in","r",stdin);
freopen("schlnet.out","w",stdout);
init();
work();
return 0;
}

----------------------------------------------------------------------------------

Network of Schools
IOI '96 Day 1 Problem 3

A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the "receiving schools"). Note that if B is in the distribution list of school A, then A does not necessarily appear in the list of school B.

You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.

PROGRAM NAME: schlnet

INPUT FORMAT

The first line of the input file contains an integer N: the number of schools in the network (2<=N<=100). The schools are identified by the first N positive integers. Each of the next N lines describes a list of receivers. The line i+1 contains the identifiers of the receivers of school i. Each list ends with a 0. An empty list contains a 0 alone in the line.

SAMPLE INPUT (file schlnet.in)

5 2 4 3 0 4 5 0 0 0 1 0 

OUTPUT FORMAT

Your program should write two lines to the output file. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.

SAMPLE OUTPUT (file schlnet.out)

1 2

[IOI1996] USACO Section 5.3 Network of Schools(强连通分量)的更多相关文章

  1. Network of Schools(强连通分量缩点(邻接表&矩阵))

    Description A number of schools are connected to a computer network. Agreements have been developed ...

  2. Network of Schools(强连通分量+缩点) (问添加几个点最少点是所有点连接+添加最少边使图强连通)

    Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13801   Accepted: 55 ...

  3. POJ 1236 Network Of Schools (强连通分量缩点求出度为0的和入度为0的分量个数)

    Network of Schools A number of schools are connected to a computer network. Agreements have been dev ...

  4. POJ1236 Network of Schools —— 强连通分量 + 缩点 + 入出度

    题目链接:http://poj.org/problem?id=1236 Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Tot ...

  5. poj-1236.network of schools(强连通分量 + 图的入度出度)

    Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 27121   Accepted: 10 ...

  6. POJ1236 Network of Schools (强连通分量,注意边界)

    A number of schools are connected to a computer network. Agreements have been developed among those ...

  7. POJ 1236 Network of Schools (强连通分量缩点求度数)

    题意: 求一个有向图中: (1)要选几个点才能把的点走遍 (2)要添加多少条边使得整个图强联通 分析: 对于问题1, 我们只要求出缩点后的图有多少个入度为0的scc就好, 因为有入度的scc可以从其他 ...

  8. POJ1236Network of Schools[强连通分量|缩点]

    Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16571   Accepted: 65 ...

  9. poj~1236 Network of Schools 强连通入门题

    一些学校连接到计算机网络.这些学校之间已经达成了协议: 每所学校都有一份分发软件的学校名单("接收学校"). 请注意,如果B在学校A的分发名单中,则A不一定出现在学校B的名单中您需 ...

随机推荐

  1. python socket理论知识

    一.socket理论: 发现一个很好的文章,一个高手写的,我也就不再做搬运工了,直接连接吧,对理论感兴趣的可以去看看! http://www.cnblogs.com/dolphinX/p/346054 ...

  2. BeanUtils\DBUtils

    BeanUtil: 需要导入 beanutil包和logging日志包 用于给对象属性赋值. setProperty与copyProperty区别: 这个问题搁置,还不会. 将map数据拷贝到对象中, ...

  3. Swift初体验(三)

    /*******************************************************************************/ // 协议 protocol Des ...

  4. 最终有SpringMvc与Struts2的对照啦

    眼下企业中使用SpringMvc的比例已经远远超过Struts2,那么两者究竟有什么差别,是非常多刚開始学习的人比較关注的问题,以下我们就来对SpringMvc和Struts2进行各方面的比較: 1. ...

  5. java 请求响应乱码

    package org.operamasks.servlet; import java.io.IOException; import java.io.PrintWriter; import java. ...

  6. [LeetCode] Search a 2D Matrix [25]

    题目 Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the fo ...

  7. md笔记——正则学习

    正则表达式 在线调试正则1 在线调试正则2 规则记录 \d 匹配一个数字字符.等价于[0-9] \D 匹配一个非数字字符.等价于[^0-9]. . 通配符,可以匹配任意字符. ? 表示量词" ...

  8. java调试工具

    jps当前用户已启动的java进程信息,信息包括进程号和简短的进程command. jstat输出指定 jvm 实例的特定统计量:统计量:-class-compiler-gc-gccapacity-g ...

  9. English - 英语中的时间表达法,这里全啦!

  10. 匹配图片src正则

    preg_match_all('#<img.*?src="([^"]*)"[^>]*>#i', $xstr, $match); /** * 获取图片sr ...