POJ1236 Network of Schools 【强连通分量Garbow】

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11180		Accepted: 4448

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

题意：给定各台电脑的连接情况，求两个问题：1、最少向网络中的几台电脑投放文件，则整个网络中的所有电脑能立刻获得该文件；2、最少要向网络中添加几条单向连接可以使得这个网络中只要投放一份文件到任意一台电脑，则所有电脑都能获得该文件。

题解：第一道Garbow题，需要注意的地方是当给定图是连通图时不需要添加任何边。

改进后的一个版本：0ms

#include <stdio.h>
#include <string.h>
#define maxn 102
#define maxm maxn * maxn

int head[maxn], sta1[maxn], sta2[maxn], scc[maxn];
int low[maxn], id, id1, id2, sec, sccNum;
bool in[maxn], out[maxn];
struct Node{
int to, next;
} E[maxm];

void addEdge(int u, int v)
{
E[id].to = v;
E[id].next = head[u];
head[u] = id++;
}

void getMap(int n)
{
memset(head, -1, sizeof(int) * (n + 1));
int i, v; id = 0;
for(i = 1; i <= n; ++i)
while(scanf("%d", &v), v)
addEdge(i, v);
}

void Garbow(int pos)
{
sta1[id1++] = sta2[id2++] = pos;
low[pos] = ++sec;
for(int i = head[pos]; i != -1; i = E[i].next){
if(!low[E[i].to]) Garbow(E[i].to);
else if(!scc[E[i].to])
while(low[E[i].to] < low[sta2[id2-1]]) --id2;
if(scc[E[i].to]) in[scc[E[i].to]] = 1;
}
if(sta2[id2-1] == pos){
++sccNum; --id2; int v;
do{
v = sta1[--id1];
scc[v] = sccNum;
} while(sta1[id1] != pos);
}
}

void solve(int n)
{
memset(low, 0, sizeof(int) * (n + 1));
memset(scc, 0, sizeof(int) * (n + 1));
memset(in, 0, sizeof(bool) * (n + 1));
memset(out, 0, sizeof(bool) * (n + 1));
int i, j, ans1 = 0, ans2 = 0; id1 = id2 = sec = sccNum = 0;
for(i = 1; i <= n; ++i)
if(!low[i]) Garbow(i);
for(i = 1; i <= n; ++i)
for(j = head[i]; j != -1; j = E[j].next)
if(scc[i] != scc[E[j].to]){
out[scc[i]] = 1; break;
}
if(sccNum != 1){
for(i = 1; i <= sccNum; ++i){
if(!out[i]) ++ans2;
if(!in[i]) ++ans1;
}
if(ans1 > ans2) ans2 = ans1;
}else ans1 = 1;
printf("%d\n%d\n", ans1, ans2);
}

int main()
{
int n;
while(scanf("%d", &n) == 1){
getMap(n);
solve(n);
}
return 0;
}

原32ms：

#include <stdio.h>
#include <string.h>
#define maxn 102
#define maxm maxn * maxn

int head[maxn], sta1[maxn], sta2[maxn], scc[maxn];
int low[maxn], id, id1, id2, sec, sccNum;
bool in[maxn], out[maxn];
struct Node{
int to, next;
} E[maxm];

void addEdge(int u, int v)
{
E[id].to = v;
E[id].next = head[u];
head[u] = id++;
}

void getMap(int n)
{
memset(head, -1, sizeof(int) * (n + 1));
int i, v; id = 0;
for(i = 1; i <= n; ++i)
while(scanf("%d", &v), v)
addEdge(i, v);
}

void Garbow(int pos)
{
sta1[id1++] = sta2[id2++] = pos;
low[pos] = ++sec;
for(int i = head[pos]; i != -1; i = E[i].next)
if(!low[E[i].to]) Garbow(E[i].to);
else if(!scc[E[i].to])
while(low[E[i].to] < low[sta2[id2-1]]) --id2;
if(sta2[id2-1] == pos){
++sccNum; --id2; int v;
do{
v = sta1[--id1];
scc[v] = sccNum;
} while(sta1[id1] != pos);
}
}

void solve(int n)
{
memset(low, 0, sizeof(int) * (n + 1));
memset(scc, 0, sizeof(int) * (n + 1));
memset(in, 0, sizeof(bool) * (n + 1));
memset(out, 0, sizeof(bool) * (n + 1));
int i, j, ans1 = 0, ans2 = 0; id1 = id2 = sec = sccNum = 0;
for(i = 1; i <= n; ++i)
if(!low[i]) Garbow(i);
for(i = 1; i <= n; ++i)
for(j = head[i]; j != -1; j = E[j].next)
if(scc[i] != scc[E[j].to]){
out[scc[i]] = 1;
in[scc[E[j].to]] = 1;
}
if(sccNum != 1){
for(i = 1; i <= sccNum; ++i){
if(!out[i]) ++ans2;
if(!in[i]) ++ans1;
}
if(ans1 > ans2) ans2 = ans1;
}else ans1 = 1;
printf("%d\n%d\n", ans1, ans2);
}

int main()
{
int n;
while(scanf("%d", &n) == 1){
getMap(n);
solve(n);
}
return 0;
}