题目大意：给出棋盘上的N个点的位置。如今问将这些点排成一行或者一列。或者对角线的最小移动步数(每一个点都仅仅能上下左右移动。一次移动一个)

解题思路：暴力＋二分图最佳完美匹配

#include <cstdio>
#include <cstring>#define N 20
#define INF 0x3f3f3f3f
#define abs(x) ((x) > 0 ?

(x) : (-(x)))

#define max(a,b)((a)>(b)?

(a):(b))

#define min(a,b)((a)<(b)?

(a):(b))

struct Node { int x, y; }node[N]; int w[N][N], left[N], Lx[N], Ly[N], slack[N]; int ans, n; bool S[N], T[N]; bool match(int i) { S[i] = true; for (int j = 1; j <= n; j++) { if (Lx[i] + Ly[j] == w[i][j] && !T[j]) { T[j] = true; if (!left[j] || match(left[j])) { left[j] = i; return true; } } else slack[j] = min(slack[j], Lx[i] + Ly[j] - w[i][j]); } return false; } void update() { int a = 1 << 30; for (int i = 1; i <= n; i++) if (S[i]) for (int j = 1; j <= n; j++) if (!T[j]) a = min(a, Lx[i] + Ly[i] - w[i][j]); for (int i = 1; i <= n; i++) { if (S[i]) Lx[i] -= a; if (T[i]) Ly[i] += a; } } void KM() { for (int i = 1; i <= n; i++) { left[i] = Ly[i] = 0; Lx[i] = -INF; for (int j = 1; j <= n; j++) Lx[i] = max(Lx[i], w[i][j]); } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) slack[j] = INF; for (;;) { for (int j = 1; j <= n; j++) S[j] = T[j] = 0; if (match(i)) break; int a = INF; for (int j = 1; j <= n; j++) if (!T[j]) a = min(a, slack[j]); for (int j = 1; j <= n; j++) { if (S[j]) Lx[j] -= a; if (T[j]) Ly[j] += a; } } } int t = 0; for (int i = 1; i <= n; i++) t += Lx[i] + Ly[i]; ans = max(ans, t); } void init() { for (int i = 1; i <= n; i++) scanf("%d%d", &node[i].x, &node[i].y); } int cas = 1; void solve() { ans = -INF; for (int j = 1; j <= n; j++) { for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) { w[i][k] = abs(node[i].x - j) + abs(node[i].y - k); w[i][k] = -w[i][k]; } } KM(); } for (int j = 1; j <= n; j++) { for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) { w[i][k] = abs(node[i].x - k) + abs(node[i].y - j); w[i][k] = -w[i][k]; } KM(); } for (int j = 1; j <= n; j++) for (int i = 1; i <= n; i++) { w[i][j] = abs(node[i].x - j) + abs(node[i].y - j); w[i][j] = -w[i][j]; } KM(); for (int j = 1; j <= n; j++) for (int i = 1; i <= n; i++) { w[i][j] = abs(node[i].x - j) + abs(node[i].y - (n - j + 1)); w[i][j] = -w[i][j]; } KM(); printf("Board %d: %d moves required.\n\n", cas++, abs(ans)); } int main() { while (scanf("%d", &n) != EOF && n) { init(); solve(); } return 0; }