大意:给你一个图,让你使用数量最小的边使得图变成强连通图。
思路:我们发现如果图是强连通的,那么每一个顶点的出度与入度必定不为0。所以我们可以将图“缩点”,然后去统计出度或者入度为0的点,取两者的最大数。(可以手推一遍)
CODE:
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
#define MAXN 50010
#define MAXM 100010
struct Edge
{
int v, next;
}edge[MAXM];
int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM];
int ind[MAXN], outd[MAXN];
int n, m;
int cnt;
int scnt, top, tot;
void init()
{
cnt = 0;
scnt = top = tot = 0;
memset(dfn, 0, sizeof(dfn));
memset(first, -1, sizeof(first));
memset(ins, 0, sizeof(ins));
memset(ind, 0, sizeof(ind));
memset(outd, 0, sizeof(outd));
}
void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u];
first[u] = cnt++;
}
void dfs(int u)
{
int v;
dfn[u] = low[u] = ++tot;
ins[u] = 1;
stack[top++] = u;
for(int e = first[u]; e != -1; e = edge[e].next)
{
v = edge[e].v;
if(!dfn[v])
{
dfs(v);
low[u] = min(low[u], low[v]);
}
else if(ins[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(dfn[u] == low[u])
{
scnt++;
do
{
v = stack[--top];
belong[v] = scnt;
ins[v] = 0;
}while(v != u);
}
}
void Tarjan()
{
for(int v = 1; v <= n; v++) if(!dfn[v])
dfs(v);
}
void solve()
{
Tarjan(); //Tarjan
if(scnt == 1) { printf("0\n"); return ;} //本身是强连通图
for(int u = 1; u <= n; u++)
{
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(belong[u] != belong[v]) ind[belong[v]]++, outd[belong[u]]++;
}
}
int max1 = 0, max2 = 0;
for(int i = 1; i <= scnt; i++)
{
if(!outd[i]) max1++;
if(!ind[i]) max2++;
}
printf("%d\n", max1 > max2 ? max1:max2);
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
init();
scanf("%d%d", &n, &m);
if(!m) { printf("%d\n", n); continue; } //4个顶点至少需要4条边
while(m--)
{
int u, v;
scanf("%d%d", &u, &v);
read_graph(u, v);
}
solve();
}
return 0;
}
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
#define MAXN 50010
#define MAXM 100010
struct Edge
{
int v, next;
}edge[MAXM];
int first[MAXN], stack[MAXN], ins[MAXN], dfn[MAXN], low[MAXN];
int belong[MAXM];
int ind[MAXN], outd[MAXN];
int n, m;
int cnt;
int scnt, top, tot;
void init()
{
cnt = 0;
scnt = top = tot = 0;
memset(dfn, 0, sizeof(dfn));
memset(first, -1, sizeof(first));
memset(ins, 0, sizeof(ins));
memset(ind, 0, sizeof(ind));
memset(outd, 0, sizeof(outd));
}
void read_graph(int u, int v)
{
edge[cnt].v = v;
edge[cnt].next = first[u];
first[u] = cnt++;
}
void dfs(int u)
{
int v;
dfn[u] = low[u] = ++tot;
ins[u] = 1;
stack[top++] = u;
for(int e = first[u]; e != -1; e = edge[e].next)
{
v = edge[e].v;
if(!dfn[v])
{
dfs(v);
low[u] = min(low[u], low[v]);
}
else if(ins[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(dfn[u] == low[u])
{
scnt++;
do
{
v = stack[--top];
belong[v] = scnt;
ins[v] = 0;
}while(v != u);
}
}
void Tarjan()
{
for(int v = 1; v <= n; v++) if(!dfn[v])
dfs(v);
}
void solve()
{
Tarjan(); //Tarjan
if(scnt == 1) { printf("0\n"); return ;} //本身是强连通图
for(int u = 1; u <= n; u++)
{
for(int e = first[u]; e != -1; e = edge[e].next)
{
int v = edge[e].v;
if(belong[u] != belong[v]) ind[belong[v]]++, outd[belong[u]]++;
}
}
int max1 = 0, max2 = 0;
for(int i = 1; i <= scnt; i++)
{
if(!outd[i]) max1++;
if(!ind[i]) max2++;
}
printf("%d\n", max1 > max2 ? max1:max2);
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
init();
scanf("%d%d", &n, &m);
if(!m) { printf("%d\n", n); continue; } //4个顶点至少需要4条边
while(m--)
{
int u, v;
scanf("%d%d", &u, &v);
read_graph(u, v);
}
solve();
}
return 0;
}