题目描述

有一个n行m列的黑白棋盘，你每次可以交换两个相邻格子（相邻是指有公共边或公共顶点）中的棋子，最终达到目标状态。要求第i行第j列的格子只能参与mi,j次交换。

输入输出格式

输入格式：

第一行包含两个整数n，m(1≤n,m≤20)。以下n行为初始状态，每行为一个包含m个字符的01串，其中0表示黑色棋子，1表示白色棋子。以下n行为目标状态，格式同初始状态。以下n行每行为一个包含m个0..9数字的字符串，表示每个格子参与交换的次数上限。

输出格式：

输出仅一行，为最小交换总次数。如果无解，输出−1。

解法：

一眼费用流系列…然而这题的处理有点诡异，首先要拆点将点的限制转化为边的限制，S→INi,OUTi→T。IN→OUT的权值是多少？分三种情况讨论：

1. 既不是起始也不是结束状态：⌊n2⌋
2. 既是起始也是结束状态：⌊n+22⌋
3. 是起始或结束状态：⌊n+12⌋

#include <bits/stdc++.h>
using namespace std;const int MAXN = 2005;
struct node {int to, next, f, c, neg;
} edge[MAXN*40];
int head[MAXN], top = 0;
void push(int i, int j, int k, int l)
{if (i*j == 0) return;++top, edge[top] = (node) {j, head[i], k, l, top+1}, head[i] = top;++top, edge[top] = (node) {i, head[j], 0, -l, top-1}, head[j] = top;
}int vis[MAXN], dis[MAXN], S = 2001, T = 2002;
queue<int> que;
int pre[MAXN], pre_edge[MAXN];
const int INF = 233333333;bool spfa()
{memset(dis, 127/3, sizeof dis);memset(pre, 0, sizeof pre);vis[S] = 1, que.push(S), dis[S] = 0;while (!que.empty()) {int tp = que.front(); que.pop(), vis[tp] = 0;for (int i = head[tp]; i; i = edge[i].next) {if (edge[i].f == 0 || dis[edge[i].to] <= dis[tp] + edge[i].c) continue;int to = edge[i].to, c = edge[i].c;dis[to] = dis[tp] + c, pre[to] = tp, pre_edge[to] = i;if (!vis[to])vis[to] = 1, que.push(to);}}return dis[T] < INF;
}int sap(int &cost)
{int ans = INF;for (int i = T; i != S; i = pre[i]) ans = min(ans, edge[pre_edge[i]].f);for (int i = T; i != S; i = pre[i]) edge[pre_edge[i]].f -= ans, edge[edge[pre_edge[i]].neg].f += ans;cost += ans*dis[T];return ans;
}int mst(int &cost)
{cost = 0;int ans = 0;while (spfa()) ans += sap(cost);return ans;
}int n, m;
int A[30][30], B[30][30], C[30][30];
char str[30];int readln(int A[30][30])
{int cnt = 0;for (int i = 1; i <= n; i++) {scanf("%s", str+1);for (int j = 1; j <= m; j++)A[i][j] = str[j]-'0', cnt += A[i][j];}return cnt;
}inline int number(int i, int j, int id)
{ return (i<=0||i>n||j<=0||j>m)?0:id*n*m+(i-1)*m+j; }int dx[] = {1,0,-1,0, 1, 1, -1, -1}, dy[] = {0,1,0,-1, -1, 1, -1, 1};
int main()
{scanf("%d%d", &n, &m);int a = readln(A);int b = readln(B); readln(C);if (a != b) {puts("-1"); return 0;}for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++) {if (A[i][j] == 1) push(S, number(i, j, 0), 1, 0);if (B[i][j] == 1) push(number(i, j, 1), T, 1, 0);if (A[i][j] == 1 && B[i][j] == 1) push(number(i, j, 0), number(i, j, 1), (C[i][j]+2)/2, 1);else if (A[i][j] == 1 || B[i][j] == 1) push(number(i, j, 0), number(i, j, 1), (C[i][j]+1)/2, 1);else push(number(i, j, 0), number(i, j, 1), C[i][j]/2, 1);for (int k = 0; k < 8; k++) push(number(i, j, 1), number(i+dx[k], j+dy[k], 0), INF, 0);}int ans = 0, cost = 0;ans = mst(cost);if (ans == a) {cout << cost-a << endl;} elseputs("-1");return 0;
}