HDU 2767 - Proving Equivalences (强连通 + 缩点)
题意
找出还要加几条边才能构成一个强连通图
思路
求出强连通分量后缩点,求出每个强连通分量的入度和出度,答案就是入度为0数目好出度为0数目的最大值。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <ctime>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-6;
const int MAXN = 2e4 + 10;
const int MOD = 1000007;
const int dir[][2] = { {1, 0}, {-1, 0}, {0, -1}, {0, 1} };
int dfs_clock, scc_cnt, pre[MAXN], lowlink[MAXN], sccno[MAXN], in[MAXN], out[MAXN];
vector<int> G[MAXN];
stack<int> st;
void DFS(int u)
{
pre[u] = lowlink[u] = ++dfs_clock;
st.push(u);
for (int i = 0; i < SZ(G[u]); 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++;
while (1)
{
int x = st.top(); st.pop();
sccno[x] = scc_cnt;
if (x == u) break;
}
}
}
void FindSCC(int v)
{
MS(pre, 0); MS(lowlink, 0); MS(sccno, 0);
dfs_clock = scc_cnt = 0;
for (int i = 1; i <= v; i++)
if (!pre[i]) DFS(i);
}
int main()
{
// ROP;
int T, i, j;
scanf("%d", &T);
while (T--)
{
MS(in, 0); MS(out, 0);
int v, e;
scanf("%d%d", &v, &e);
for (i = 0; i <= v; i++) G[i].clear();
for (i = 0; i < e; i++)
{
int a, b;
scanf("%d%d", &a, &b);
G[a].PB(b);
}
FindSCC(v);
if (scc_cnt == 1)
{
puts("0");
continue;
}
for (i = 1; i <= v; i++)
{
for (j = 0; j < SZ(G[i]); j++)
{
if (sccno[i] != sccno[G[i][j]])
{
out[sccno[i]]++;
in[sccno[G[i][j]]]++;
}
}
}
int inAns = 0, outAns = 0;
for (i = 1; i <= scc_cnt; i++)
{
if (in[i] == 0) inAns++;
if (out[i] == 0) outAns++;
}
printf("%d\n", max(inAns, outAns));
}
return 0;
}