UVA - 1045 The Great Wall Game(二分图最佳完美匹配)

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;
}