Blog of RuSun

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

P2055 [ZJOI2009]假期的宿舍

P2055 [ZJOI2009]假期的宿舍

根据题意二分图匹配建边即可,注意细节的处理。

查看代码
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#include <iostream>
#include <cstdio>
#include <queue>
#include <climits>
#define INF INT_MAX
using namespace std;
const int N = 110, M = 5.2e3 + 10;
int n, st, ed, tot, d[N], cur[N];
int idx, hd[N], nxt[M], edg[M], wt[M];
bool people[N], bed[N];
bool bfs()
{
for (int i = st; i <= ed; i++)
d[i] = -1;
d[st] = 0;
cur[st] = hd[st];
queue <int> q;
q.push(st);
while (!q.empty())
{
int t = q.front();
q.pop();
for (int i = hd[t]; ~i; i = nxt[i])
if (d[edg[i]] == -1 && wt[i])
{
cur[edg[i]] = hd[edg[i]];
d[edg[i]] = d[t] + 1;
if (edg[i] == ed)
return true;
q.push(edg[i]);
}
}
return false;
}
int find(int x, int limit)
{
if (x == ed)
return limit;
int res = 0;
for (int i = cur[x]; ~i && res < limit; i = nxt[i])
{
cur[x] = i;
if (d[edg[i]] == d[x] + 1 && wt[i])
{
int t = find(edg[i], min(wt[i], limit - res));
if (!t)
d[edg[i]] = -1;
wt[i] -= t;
wt[i ^ 1] += t;
res += t;
}
}
return res;
}
int dinic()
{
int res = 0, flow;
while (bfs())
while (flow = find(st, INF))
res += flow;
return res;
}
void add(int x, int y, int z)
{
nxt[++idx] = hd[x];
hd[x] = idx;
edg[idx] = y;
wt[idx] = z;
}
int main()
{
int T;
cin >> T;
while (T--)
{
cin >> n;
st = 0;
ed = n << 1 | 1;
idx = -1;
tot = 0;
for (int i = st; i <= ed; i++)
hd[i] = -1;
for (int i = 1; i <= n; i++)
cin >> bed[i];
for (int i = 1; i <= n; i++)
{
cin >> people[i];
if (!bed[i])
people[i] = true;
else
people[i] = people[i] ^ 1;
tot += people[i];
}
for (int i = 1; i <= n; i++)
if (people[i])
{
add(st, i, 1);
add(i, st, 0);
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
{
bool a;
cin >> a;
if ((a || i == j) && people[i] && bed[j])
{
add(i, j + n, 1);
add(j + n, i, 0);
}
}
for (int i = 1; i <= n; i++)
if (bed[i])
{
add(i + n, ed, 1);
add(ed, i + n, 0);
}
if (dinic() == tot)
cout << "^_^\n";
else
cout << "T_T\n";
}
return 0;
}