HDU 2767 Proving Equivalences （tarjan scc）

题意:

求给定的图最少需要添加多少条边,整个图是强联通的

分析:

分析强联通图的性质可知,所有点的入度和出度都至少为1

如果要让此图连通,只要在scc缩点之后,对于新图中,求出根数(入度为0)和叶子数(出度为0),把根和叶子连起来,图就连通了

答案显然是ans=max(根数,叶子数)

需要特判整个图已经强联通时ans=0

代码:

//
//  Created by TaoSama on 2015-11-21
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 5e4 + 10;

int n, m;
struct Edge {
int v, nxt;
} edge[M];
int head
, cnt;

void addEdge(int u, int v) {
edge[cnt] = (Edge) {v, head[u]};
head[u] = cnt++;
}

int dfn
, low
, in
, id
, scc, dfsNum;
int stk
, top;

void tarjan(int u) {
dfn[u] = low[u] = ++dfsNum;
stk[++top] = u;
in[u] = true;
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if(in[v]) low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u]) {
++scc;
while(true) {
int v = stk[top--];
in[v] = false;
id[v] = scc;
if(v == u) break;
}
}
}

void init() {
scc = dfsNum = 0;
memset(dfn, 0, sizeof dfn);
}

int inDeg
, outDeg
;

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
cnt = 0; memset(head, -1, sizeof head);
while(m--) {
int u, v; scanf("%d%d", &u, &v);
addEdge(u, v);
}
init();
for(int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i);
if(scc == 1) {puts("0"); continue;} //已经连通

memset(inDeg, 0, sizeof inDeg);
memset(outDeg, 0, sizeof outDeg);
for(int j = 1; j <= n; ++j) {
int u = id[j];
for(int i = head[j]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
v = id[v];
if(u == v) continue;
++outDeg[u]; ++inDeg[v];
}
}

int root = 0, leaf = 0;
for(int i = 1; i <= scc; ++i) {
if(inDeg[i] == 0) ++root;
if(outDeg[i] == 0) ++leaf;
}
printf("%d\n", max(root, leaf));
}
return 0;
}