版权声明: https://blog.csdn.net/moon_sky1999/article/details/81539741
题目来源:http://poj.org/problem?id=1236
最开始做的时候理解错题意了QWQ,WA了几发。
题目大意:
给定一个有向图,第一问是选择部分的节点,以这些节点为起点可以遍历整个图,最少要选几个节点。
第二问是问添加最少多少条边,使得以任意节点为起点,均可以遍历整个图。
思路:
先用tarjan求出整个图的所有强连通分量,对每个连通分量进行缩点,处理出缩点后的图。
第一问即是找出缩点后的图中所有入度为0的点。
第二问是添加最少多少条边让这个图变为强连通图。也就是max{入度为0的点,出度为0的点}。(首尾相连)
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <iostream>
#define ll long long
using namespace std;
int n, cur, top, color, dfn[111], s[111], low[111], c[111], num[111], du[111], dd[111];
bool g[111][111], vis[111], e[111][111];
void tarjan(int u) {
dfn[u] = low[u] = ++cur;
vis[u] = 1;
s[++top] = u;
for (int i = 1; i <= n; ++i) {
if (g[u][i]) {
if (dfn[i] == 0)tarjan(i);
if (vis[i] && low[u] > low[i])low[u] = low[i];
}
}
if (dfn[u] == low[u]) {
color++;
while (s[top + 1] != u) {
vis[s[top]] = 0;
c[s[top]] = color;
num[color]++;
top--;
}
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
while (cin >> n) {
int x;
memset(g, 0, sizeof(g));
memset(c, 0, sizeof(c));
memset(num, 0, sizeof(num));
memset(s, 0, sizeof(s));
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
memset(vis, 0, sizeof(vis));
memset(du, 0, sizeof(du));
memset(dd, 0, sizeof(dd));
for (int i = 1; i <= n; ++i) {
while (cin >> x) {
if (x == 0)break;
g[i][x] = 1;
}
}
cur = top = color = 0;
for (int i = 1; i <= n; ++i) {
if (dfn[i] == 0) {
tarjan(i);
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (g[i][j] && c[i] != c[j]) {
e[c[i]][c[j]] = 1;
}
}
}
for (int i = 1; i <= color; ++i) {
for (int j = 1; j <= color; ++j) {
if (e[i][j]) {
du[j]++;
dd[i]++;
}
}
}
int ans = 0, tot = 0;
for (int i = 1; i <= color; ++i) {
if (du[i] == 0)
ans++;
}
cout << ans << endl;
for (int i = 1; i <= color; ++i) {
if (dd[i] == 0)
tot++;
}
if (color == 1)cout << 0 << endl;
else cout << max(tot, ans) << endl;
}
return 0;
}