本文主要是介绍UVA 1045 - The Great Wall Game（二分图完美匹配），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！
UVA 1045 - The Great Wall Game
题目链接
题意：给定一个n*n的棋盘，有n个棋子在上面，现在要移动棋子，每一步代价是1，现在要把棋子移动到一行，一列，或者在主副对角线上，问最小代价
思路：二分图完美匹配，枚举每种情况，建边，边权为曼哈顿距离，然后km算法做完美匹配算出值即可，由于要求最小值所以边权传负数，这样做出来的值的负就是答案
代码：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;const int MAXNODE = 105;typedef int Type;
const Type INF = 0x3f3f3f3f;struct KM {int n;Type g[MAXNODE][MAXNODE];Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];int left[MAXNODE];bool S[MAXNODE], T[MAXNODE];void init(int n) {this->n = n;}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 < n; 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;return true;}} else slack[j] = min(slack[j], tmp);}return false;}void update() {Type a = INF;for (int i = 0; i < n; i++)if (!T[i]) a = min(a, slack[i]);for (int i = 0; i < n; i++) {if (S[i]) Lx[i] -= a;if (T[i]) Ly[i] += a;}}Type km() {for (int i = 0; i < n; i++) {left[i] = -1;Lx[i] = -INF; Ly[i] = 0;for (int j = 0; j < n; j++)Lx[i] = max(Lx[i], g[i][j]);}for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) slack[j] = INF;while (1) {for (int j = 0; j < n; j++) S[j] = T[j] = false;if (dfs(i)) break;else update();}}Type ans = 0;for (int i = 0; i < n; i++)ans += g[left[i]][i];return ans;}
} gao;const int N = 20;int n, x[N], y[N];int dis(int x1, int y1, int x2, int y2) {return abs(x1 - x2) + abs(y1 - y2);
}int main() {int cas = 0;while (~scanf("%d", &n) && n) {gao.init(n);for (int i = 0; i < n; i++) {scanf("%d%d", &x[i], &y[i]);x[i]--; y[i]--;}int ans = -1000;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {for (int k = 0; k < n; k++) {gao.add_Edge(j, k, -dis(x[j], y[j], i, k));}}ans = max(ans, gao.km());}for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {for (int k = 0; k < n; k++) {gao.add_Edge(j, k, -dis(x[j], y[j], k, i));}}ans = max(ans, gao.km());}for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)gao.add_Edge(i, j, -dis(x[i], y[i], j, j));ans = max(ans, gao.km());for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)gao.add_Edge(i, j, -dis(x[i], y[i], n - j - 1, j));ans = max(ans, gao.km());printf("Board %d: %d moves required.\n\n", ++cas, -ans);}return 0;
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