一名苦逼的OIer,想成为ACMer

Iowa_Battleship

BZOJ1433或洛谷2055 [ZJOI2009]假期的宿舍

BZOJ原题链接

洛谷原题链接

对于每个需要床位的人向他能睡的床连边,然后就是二分图最大匹配模板了。
这里用匈牙利算法。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 55;
const int M = 1e4 + 10;
int fi[M], di[M], ne[M], mtc[M], l;
bool v[M], h[N], sc[N];
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
inline void add(int x, int y)
{
	di[++l] = y;
	ne[l] = fi[x];
	fi[x] = l;
}
bool dfs(int x)
{
	int i, y;
	for (i = fi[x]; i; i = ne[i])
		if (!v[y = di[i]])
		{
			v[y] = 1;
			if (!mtc[y] || dfs(mtc[y]))
			{
				mtc[y] = x;
				return true;
			}
		}
	return false;
}
int main()
{
	int i, j, n, t;
	t = re();
	while (t--)
	{
		n = re();
		l = 0;
		memset(fi, 0, sizeof(fi));
		memset(mtc, 0, sizeof(mtc));
		for (i = 1; i <= n; i++)
			sc[i] = re();
		for (i = 1; i <= n; i++)
			sc[i] ? h[i] = re() : h[i] = re() ? 0 : 0;
		int k = 0;
		for (i = 1; i <= n; i++)
			if (!sc[i] || (sc[i] && !h[i]))
				k++;
		for (i = 1; i <= n; i++)
		{
			if (!h[i] && sc[i])
				add(i, i + n);
			for (j = 1; j <= n; j++)
				if (re() && !h[i] && sc[j])
					add(i, j + n);
		}
		int s = 0;
		for (i = 1; i <= n; i++)
		{
			memset(v, 0, sizeof(v));
			if (dfs(i))
				s++;
		}
		s ^ k ? printf("T_T\n") : printf("^_^\n");
	}
	return 0;
}

posted on 2018-11-04 21:26  Iowa_Battleship  阅读(48)  评论(0编辑  收藏

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