C - Network of Schools
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d
& %I64u
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
这个题最大的难点在于任务二，任务二的意思就是添加最少的边使一个有向图变成强连通图，有一个定理是max（n,m)其中n是出度为0的点的个数，m是入度为0的点的个数，当然，如果这个图是强连通图的话，就需要讨论了，这时答案就是0了。将第二个问题解决掉，这个题就是一个模板题了，但是我现在仍没有证明这个命题的正确性！有大神说显然，我觉得应该可以证出来，希望看到这个博客的大神可以帮帮忙，谢谢。
#include<stdio.h>
#include<stack>
#include<string.h>
#include<algorithm>
using namespace std;
int dfn[120];
int belong[120],bnt,instack[120];
int index,out[120],in[120],low[120];
int map[120][120];
stack<int>S;
void tarjan(int i){
dfn[i] = low[i] = ++index;
S.push(i);
instack[i] = 1;
for(int j = 1;j<=map[i][0];j++){
int k = map[i][j];
if(!dfn[k]){
tarjan(k);
low[i] = min(low[i],low[k]);
}
else if(instack[k])
low[i] = min(low[i],dfn[k]);
}
if(low[i] == dfn[i]){
bnt++;
int j;
do{
j = S.top();
S.pop();
instack[j] = 0;
belong[j] = bnt;
}while(i!=j);
}
}
int main(){
int n,m,ans,ans1;
while(~scanf("%d",&n)){
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(instack,0,sizeof(instack));
memset(out,0,sizeof(out));
memset(in,0,sizeof(in));
memset(map,0,sizeof(map));
bnt = index = 0;
ans = ans1 = 0;
for(int i=1;i<=n;i++){
while(scanf("%d",&m),m)
if(m)map[i][++map[i][0]] = m;
}
for(int i=1;i<=n;i++)
if(!dfn[i])tarjan(i);
for(int i=1;i<=n;i++){
for(int j = 1;j<=map[i][0];j++){
if(belong[i]!=belong[map[i][j]]){
out[belong[i]]++;
in[belong[map[i][j]]]++;
}
}
}
for(int i=1;i<=bnt;i++){
if(out[i]==0)ans++;
if(in[i]==0)ans1++;
}
printf("%d\n",ans1);
if(bnt ==1)printf("0\n");
else printf("%d\n",max(ans1,ans));
}
}
tarjan+缩点+强连通定理