HDU 3995 Special Fish

题目链接

题意：一些鱼，每只鱼都有一个权值，给一个矩阵，如果mat[i][j] = 1表示i会攻击j，每只鱼可以攻击一次和被攻击一次，每次攻击可以得到权值为val[i]^val[j]，问最大能得到多少权值

思路：KM最大匹配，每个鱼拆成攻击和被攻击两边，然后连边跑KM最大匹配即可

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int MAXNODE = 105;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct KM {
	int n, m;
	Type g[MAXNODE][MAXNODE];
	Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];
	int left[MAXNODE], right[MAXNODE];
	bool S[MAXNODE], T[MAXNODE];

	void init(int n, int m) {
		this->n = n;
		this->m = m;
		memset(g, 0, sizeof(g));
	}

	void add_Edge(int u, int v, Type val) {
		g[u][v] = val;
	}

	bool dfs(int i) {
		S[i] = true;
		for (int j = 0; j < m; j++) {
			if (T[j]) continue;
			Type tmp = Lx[i] + Ly[j] - g[i][j];
			if (!tmp) {
				T[j] = true;
				if (left[j] == -1 || dfs(left[j])) {
					left[j] = i;
					right[i] = j;
					return true;
				}
			} else slack[j] = min(slack[j], tmp);
		}
		return false;
	}

	void update() {
		Type a = INF;
		for (int i = 0; i < m; i++)
			if (!T[i]) a = min(a, slack[i]);
		for (int i = 0; i < n; i++)
			if (S[i]) Lx[i] -= a;
		for (int i = 0; i < m; i++)
			if (T[i]) Ly[i] += a;
	}

	Type km() {
		memset(left, -1, sizeof(left));
		memset(right, -1, sizeof(right));
		memset(Ly, 0, sizeof(Ly));
		for (int i = 0; i < n; i++) {
			Lx[i] = -INF;
			for (int j = 0; j < m; j++)
				Lx[i] = max(Lx[i], g[i][j]);
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) slack[j] = INF;
			while (1) {
				memset(S, false, sizeof(S));
				memset(T, false, sizeof(T));
				if (dfs(i)) break;
				else update();
			}
		}
		Type ans = 0;
		for (int i = 0; i < n; i++)
			ans += g[i][right[i]];
		return ans;
	}
} gao;

const int N = 105;

int n, val[N];
char str[N];

int main() {
	while (~scanf("%d", &n) && n) {
		gao.init(n, n);
		for (int i = 0; i < n; i++) scanf("%d", &val[i]);
		for (int i = 0; i < n; i++) {
			scanf("%s", str);
			for (int j = 0; j < n; j++) {
				if (str[j] == '1') {
					gao.add_Edge(i, j, val[i]^val[j]);
				}
			}
		}
		printf("%d\n", gao.km());
	}
	return 0;
}