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BZOJ 2668

題解

同時分別限制流入和流出次數，所以把一個點拆成三個：入點in(x)、中間點mi(x)、出點ou(x)。

如果一個格子x在初始狀態是黑點，則連(S, mi(x), 1, 0)
如果x在目標狀態是黑點，則連(mi(x), T, 1, 0)
設x的交換次數限制是w
如果x在兩種狀態中顏色相同，則連(in(x), mi(x), w / 2, 0), (mi(x), ou(x), w / 2, 0)
如果x只在初始狀態為黑色，則連(in(x), mi(x), w / 2, 0), (mi(x), ou(x), (w + 1) / 2, 0)
如果x只在目標狀態為黑色，則連(in(x), mi(x), (w + 1) / 2, 0), (mi(x), ou(x), w / 2, 0)

然後最小費用最大流即可。

#include <queue>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
#define space putchar(' ')
#define enter putchar('\n')
typedef long long ll;
using
 
 namespace std;
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10
 
 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}

const int N = 1234, M = 1000005, INF = 0x3f3f3f3f;
const int dx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};
int n, m, tot, maxflow, mincost, src = 1, des = 2, id[23][23][3], b1, b2;
int ecnt = 1, adj[N], pre[N], dis[N], nxt[M], go[M], cap[M], cost[M];
char st[23][23], ed[23][23], cp[23][23];

void _add(int u, int v, int w, int c){
    go[++ecnt] = v;
    nxt[ecnt] = adj[u];
    adj[u] = ecnt;
    cap[ecnt] = w;
    cost[ecnt] = c;
}
void add(int u, int v, int w, int c){
    _add(u, v, w, c);
    _add(v, u, 0, -c);
}
bool spfa(){
    queue <int> que;
    static bool inq[N] = {0};
    for(int i = 1; i <= tot; i++)
        dis[i] = INF, pre[i] = 0;
    dis[src] = 0, que.push(src), inq[src] = 1;
    while(!que.empty()){
        int u = que.front();
        que.pop(), inq[u] = 0;
        for(int e = adj[u], v; e; e = nxt[e]){
            if(cap[e] && dis[u] + cost[e] < dis[v = go[e]]){
                dis[v] = dis[u] + cost[e], pre[v] = e;
                if(!inq[v]) que.push(v), inq[v] = 1;
            }
        }
    }
    return dis[des] < INF;
}
void mcmf(){
    while(spfa()){
        int delta = INF;
        for(int e = pre[des]; e; e = pre[go[e ^ 1]])
            delta = min(delta, cap[e]);
        for(int e = pre[des]; e; e = pre[go[e ^ 1]])
            cap[e] -= delta, cap[e ^ 1] += delta;
        maxflow += delta;
        mincost += delta * dis[des];
    }
}
bool legal(int x, int y){
    return x > 0 && y > 0 && x <= n && y <= m;
}

int main(){

    read(n), read(m), tot = 3 * n * m + 2;
    for(int i = 1, cnt = 2; i <= n; i++)
        for(int j = 1; j <= m; j++)
            id[i][j][0] = ++cnt, id[i][j][1] = ++cnt, id[i][j][2] = ++cnt;
    for(int i = 1; i <= n; i++) scanf("%s", st[i] + 1);
    for(int i = 1; i <= n; i++) scanf("%s", ed[i] + 1);
    for(int i = 1; i <= n; i++) scanf("%s", cp[i] + 1);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++){
            int w = cp[i][j] - '0';
            if(st[i][j] == '1') b1++, add(src, id[i][j][0], 1, 0);
            if(ed[i][j] == '1') b2++, add(id[i][j][0], des, 1, 0);
            if(st[i][j] == ed[i][j]){
                add(id[i][j][1], id[i][j][0], w / 2, 0);
                add(id[i][j][0], id[i][j][2], w / 2, 0);
            }
            else if(st[i][j] == '1'){
                add(id[i][j][1], id[i][j][0], w / 2, 0);
                add(id[i][j][0], id[i][j][2], (w + 1) / 2, 0);
            }
            else if(ed[i][j] == '1'){
                add(id[i][j][1], id[i][j][0], (w + 1) / 2, 0);
                add(id[i][j][0], id[i][j][2], w / 2, 0);
            }
            for(int d = 0, x, y; d < 8; d++)
                if(legal(x = i + dx[d], y = j + dy[d]))
                    add(id[i][j][2], id[x][y][1], INF, 1);
        }
    mcmf();
    if(maxflow < max(b1, b2)) puts("-1");
    else write(mincost), enter;
    
    return 0;
}

BZOJ 2668 [cqoi2012]交換棋子 | 最小費用最大流