LA 3276 The Great Wall Game(二分图最佳完美匹配)

题意：n * n的棋盘上有n个棋子，现在要求移动这些棋子，使这些棋子在同一行或者同一列，每次只能移动一步，问最少需要多少次移动能满足要求。

思路：因为每个棋子到每一条直线上的n个格子都有n个移动，每次将每个棋子和直线上的n个格子连线，边权为移动的步数的负数，然后求一下完美匹配就可以了。

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
#define bug printf("****\n");
const int maxn = 20;
using namespace std;

int w[maxn][maxn], n;
int lx[maxn], ly[maxn];
int from[maxn], kase = 1;
bool S[maxn], T[maxn];
int x[maxn], y[maxn];

bool match(int i) {
S[i] = true;
for(int j = 1; j <= n; j++) {
if(lx[i] + ly[j] != w[i][j] || T[j]) continue;
T[j] = true;
if(!from[j] || match(from[j])) {
from[j] = i;
return true;
}
}
return false;
}

void update() {
int a = 1 << 30;
for(int i = 1; i <= n; i++) {
if(!S[i]) continue;
for(int j = 1; j <= n; j++) {
if(T[j]) continue;
a = min(a, lx[i] + ly[j] - 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++) {
from[i] = lx[i] = ly[i] = 0;
for(int j = 1; j <= n; j++) {
lx[i] = max(lx[i], w[i][j]);
}
}
for(int i = 1; i <= n; i++) {
while(1) {
memset(S, 0, sizeof S);
memset(T, 0, sizeof T);
int cc = match(i);

if(cc) break;
else update();
}
}
}

int cost() {
int sum = 0;
for(int i = 1; i <= n; i++) {
sum += w[from[i]][i];
}
return sum;
}

int main() {
while(scanf("%d", &n) && n) {
int ans = -1e8;
for(int i = 1; i <= n; i++) {
scanf("%d %d", &x[i], &y[i]);
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
for(int k = 1; k <= n; k++) {
w[j][k] = abs(i - x[k]) + abs(j - y[k]);
w[j][k] = -w[j][k];
}
}
KM(); ans = max(ans, cost());
for(int j = 1; j <= n; j++) {
for(int k = 1; k <= n; k++) {
w[j][k] = abs(j - x[k]) + abs(i - y[k]);
w[j][k] = -w[j][k];
}
}
KM(); ans = max(ans, cost());
}
for(int i = 1; i <= n; i++) {
for(int k = 1; k <= n; k++) {
w[i][k] = abs(i - x[k]) + abs(i - y[k]);
w[i][k] = -w[i][k];
}
}
KM(); ans = max(ans, cost());
for(int i = 1; i <= n; i++) {
for(int k = 1; k <= n; k++) {
w[i][k] = abs(i - x[k]) + abs(n - i + 1 - y[k]);
w[i][k] = -w[i][k];
}
}
KM(); ans = max(ans, cost());
printf("Board %d: %d moves required.\n\n", kase++, -ans);
}
return 0;
}