Network of Schools
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 1236
Description
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.
Input
The first line 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.
Output
Your program should write two lines to the standard output. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.
Sample Input
5 2 4 3 0 4 5 0 0 0 1 0
Sample Output
1 2 题目大意：给你n个学校，每个学校有一个接收表，表示城市i向表内城市都可以发送文件。现在要发送文件给所有学校，问题A：问你至少发给几个学校就可以通过学校间的发送，让所有学校都能收到文件。问题B：问你发个任意一个学校，要让所有学校都能收到文件，需要最少加多少条有向边。 解题思路：问题A，其实就是让求对于强连通分量缩点后，有多少个缩点的入度为0。问题B：其实就让你求至少加入多少条有向边，可以让原图形成强连通图。问题B，对于缩点，我们统计入度和出度。对于入度为0的，我们理论上应该给它加一条入边，对于出度为0的，我们应该给它加一条出边。但是要求最少加边数，那么我们就可以让入度为0的和出度为0的各取所需，即从出度为0的缩点向入度为0的缩点连一条边。对于最后可能会剩下的一个点，让他另外加一条边即可。那么我们的答案就很明显了，即max(入度为0的缩点个数，出度为0的缩点个数)。
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<vector>
#include<stack>
#include<iostream>
using namespace std;
const int maxn = 210;
vector<int>G[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
stack<int>S;
int indeg[maxn], outdeg[maxn];
void dfs(int u){
pre[u] = lowlink[u] = ++dfs_clock;
S.push(u);
for(int i = 0; i < G[u].size(); i++){
int v = G[u][i];
if(!pre[v]){
dfs(v);
lowlink[u] = min(lowlink[u],lowlink[v]);
}else if(!sccno[v]){
lowlink[u] = min(lowlink[u],pre[v]);
}
}
if(lowlink[u] == pre[u]){
scc_cnt++;
for(;;){
int x = S.top(); S.pop();
sccno[x] = scc_cnt;
if(x == u) break;
}
}
}
void find_scc(int n){
dfs_clock = scc_cnt = 0;
memset(sccno,0,sizeof(sccno));
memset(pre,0,sizeof(pre));
for(int i = 1; i <= n; i++){
if(!pre[i]) dfs(i);
}
}
int main(){
int n;
scanf("%d",&n);
for(int i = 1; i <= n; i++){
int v ;
for(;;){
scanf("%d",&v);
if(v == 0){
break;
}
G[i].push_back(v);
}
}
find_scc(n);
if(scc_cnt == 1){
puts("1");
puts("0");
}else{
int sccout, sccin;
for(int i = 1; i <= n; ++i){
for(int j = 0; j < G[i].size(); j++){
int v = G[i][j];
sccout = sccno[i];
sccin = sccno[v];
if(sccout == sccin){
continue;
}
outdeg[sccout]++;
indeg[sccin]++;
}
}
int ans1 = 0, ans2 = 0;
for(int i = 1; i <= scc_cnt; i++){
if(indeg[i] == 0){
ans1++;
}
if(outdeg[i] == 0){
ans2++;
}
}
printf("%d\n%d\n",ans1,max(ans1,ans2));
}
return 0;
}