本文主要是介绍HDU 3995 Special Fish（KM最大匹配），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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;
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