UVA 1045 - The Great Wall Game（二分图完美匹配）

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
, y
;

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;
}