[CQOI2012]交换棋子

解题报告

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$m_{i,j}$次交换。

输入输出格式

输入格式：

第一行包含两个整数$n，m(1\le n, m\le 20)$。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，$S\rightarrow IN_i, OUT_i\rightarrow T$。$IN\rightarrow OUT$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：$\lfloor \frac n 2\rfloor$
2. 既是起始也是结束状态：$\lfloor \frac {n+2} 2\rfloor$
3. 是起始或结束状态：$\lfloor \frac {n+1} 2\rfloor$

#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;
    } else
        puts("-1");
    return 0;
}