题意:
给定一个有向图,求:
1) 至少要选几个顶点,才能做到从这些顶点出发,可以到达全部顶点
2) 至少要加多少条边,才能使得从任何一个顶点出发,都能到达全部顶点
思路:
1、求出所有强连通分量;
2、每个强连通分量缩成一点,则形成一个有向无环图DAG;
3、DAG上面有多少个入度为0的顶点,问题1的答案就是多少;
4、假定有 n 个入度为0的点,m个出度为0的点,max(m,n)就是第二个问题的解。
反思:
有向无环图中所有入度不为0的点,一定可以由某个入度为0的点出发可达。
代码:
#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;
const int MAXN = 100 + 10;
vector<int> graph[MAXN];
int n;
int dfn[MAXN], inStack[MAXN], st[MAXN], low[MAXN], index = 0, top = 0;
int belong[MAXN], cnt = 0;
int indegree[MAXN], outdegree[MAXN];
void tarjan(int u)
{
dfn[u] = low[u] = ++index;
inStack[u] = 1;
st[top++] = u;
for(int i = 0; i < graph[u].size(); i++)
{
int v = graph[u][i];
if(!inStack[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
}
if(inStack[v] == 1)
{
low[u] = min(low[u], dfn[v]);
}
}
if(dfn[u] == low[u])
{
int x;
++cnt;
do
{
x = st[--top];
inStack[x] = 2;
belong[x] = cnt;
}
while(x != u);
}
}
void solve()
{
for(int u = 1; u <= n; u++)
{
for(int i = 0; i < graph[u].size(); i++)
{
int v = graph[u][i];
if(belong[u] != belong[v])
{
outdegree[belong[u]]++;
indegree[belong[v]]++;
}
}
}
int amt1 = 0, amt2 = 0;//amt1:indegree = 0 amt2:outdegree = 0
for(int i = 1; i <= cnt; i++)
{
if(outdegree[i] == 0) amt2++;
if(indegree[i] == 0) amt1++;
}
printf("%d\n", amt1);
if(cnt == 1) printf("0\n");
else printf("%d\n", max(amt1, amt2));
}
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
int x;
while(scanf("%d", &x) && x)
{
graph[i].push_back(x);
}
}
for(int i = 1; i <= n; i++)
{
if(inStack[i] == 0) tarjan(i);
}
solve();
return 0;
}
/*
4
0
1 3 0
0
3 0
3
2 0
3 0
1 0
*/