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http://acm.hdu.edu.cn/showproblem.php?pid=3836
To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.
The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.
For each case, output a single integer: the minimum steps needed.
4 0
3 2
1 2
1 3
4
3
题目大意：给你一张有向图要求最少加多少条边时该图变成强连通图。
Tarjan缩点。。
#include<bits/stdc++.h>
using namespace std;
const int N = 20100;
struct Tarjan_scc {
stack<int> s;
bool instack[N];
struct edge { int to, next; }G[N * 3];
int idx, scc, tot, in[N], out[N], dfn[N], low[N], head[N], sccnum[N];
inline void init(int n) {
idx = scc = tot = 0;
while (!s.empty()) s.pop();
for (int i = 0; i < n + 2; i++) {
head[i] = -1;
instack[i] = false;
in[i] = out[i] = dfn[i] = low[i] = sccnum[i] = 0;
}
}
inline void add_edge(int u, int v) {
G[tot].to = v, G[tot].next = head[u], head[u] = tot++;
}
inline void built(int m) {
int u, v;
while (m--) {
scanf("%d %d", &u, &v);
add_edge(u, v);
}
}
inline void tarjan(int u) {
dfn[u] = low[u] = ++idx;
instack[u] = true;
s.push(u);
for (int i = head[u]; ~i; i = G[i].next) {
int &v = G[i].to;
if (!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if (instack[v] && dfn[v] < low[u]) {
low[u] = dfn[v];
}
}
if (dfn[u] == low[u]) {
int v = 0;
scc++;
do {
v = s.top(); s.pop();
instack[v] = false;
sccnum[v] = scc;
} while (u != v);
}
}
inline void solve(int n, int m) {
init(n);
built(m);
for (int i = 1; i <= n; i++) {
if (!dfn[i]) tarjan(i);
}
int x1 = 0, x2 = 0;
for (int u = 1; u <= n; u++) {
for (int i = head[u]; ~i; i = G[i].next) {
int v = G[i].to;
if (sccnum[u] != sccnum[v]) {
in[sccnum[v]]++;
out[sccnum[u]]++;
}
}
}
for (int i = 1; i <= scc; i++) {
if (!in[i]) x1++;
if (!out[i]) x2++;
}
printf("%d\n", 1 == scc ? 0 : max(x1, x2));
}
}go;
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int n, m;
while (~scanf("%d %d", &n, &m)) {
go.solve(n, m);
}
return 0;
}
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原文地址：http://www.cnblogs.com/GadyPu/p/5003447.html
