Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P1407 [国家集训队]稳定婚姻

P1407 [国家集训队]稳定婚姻

应该二分图匹配可以过的,匈牙利中,对于一对夫妻,只用做一次增广即可,但是 dinic 需要将整个图做一次,所以过不了。

正解是 tarjan ,如图,h & w 表示是夫妻关系,lovers 表示使情人关系,如果如图形成了一个环,那么每个人都可以选择一个就近的情人。

查看代码
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#include <iostream>
#include <cstdio>
#include <map>
#include <stack>
#include <string>
using namespace std;
const int N = 8e3 + 10, M = 3e4 + 10;
int n, m;
bool vis[N];
stack <int> stk;
map <string, int> name;
int idx = -1, hd[N], nxt[M], edg[M];
int cnt, stmp, id[N], dfn[N], low[N];
void tarjan (int x)
{
dfn[x] = low[x] = ++stmp;
stk.push(x);
vis[x] = true;
for (int i = hd[x]; ~i; i = nxt[i])
if (!dfn[edg[i]])
{
tarjan(edg[i]);
low[x] = min(low[x], low[edg[i]]);
}
else if (vis[edg[i]])
low[x] = min(low[x], dfn[edg[i]]);
if (dfn[x] == low[x])
{
cnt++;
int y;
do
{
y = stk.top();
stk.pop();
vis[y] = false;
id[y] = cnt;
} while (y != x);
}
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main ()
{
string a, b;
cin >> n;
for (int i = 0; i < n << 1; i++)
hd[i] = -1;
for (int i = 0; i < n; i++)
{
cin >> a >> b;
name[a] = i << 1;
name[b] = i << 1 | 1;
add(i << 1, i << 1 | 1);
}
cin >> m;
for (int i = 0; i < m; i++)
{
cin >> a >> b;
add(name[b], name[a]);
}
for (int i = 0; i < n << 1; i++)
if (!dfn[i])
tarjan(i);
for (int i = 0; i < n; i++)
if (id[i << 1] == id[i << 1 | 1])
cout << "Unsafe" << endl;
else
cout << "Safe" << endl;
return 0;
}