【题解】CQOI2012交换棋子

感受到网络流的强大了……这道题目的关键在于：

　　前后颜色不变的，流入流出的次数相等；原本是黑色的最后变成了白色，流出比流入次数多1；原本是白色最后变成黑色，流入比流出次数多一。所以我们将每一点拆成3个点，分别代表流入点，原点与流出点。最开始为黑色的点与源点连流量为1，费用为0的边，最后为黑色的点与汇点连流量为1，费用为0的边。

#include<bits/stdc++.h>
using namespace std;
#define maxn 300
#define maxm 8000
#define INF 99999
int n, m, size, tem, a[maxn][maxn], b[maxn][maxn], c[maxn][maxn], head[maxm];
int cnp, fans, ans, cost, dis[maxm], pre[maxm], flow[maxm];
int dx[10] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dy[10] = {0, 1, -1, 0, 1, -1, 1, 0, -1};
int s = 0, t; 
bool vis[maxm];
deque <int> q;

struct edge
{
    int to, last, f, c;
}E[maxn * 400];

void add(int u, int v, int f, int c)
{
    E[cnp].to = v, E[cnp].last = head[u], E[cnp].f = f, E[cnp].c = c; head[u] = cnp ++;
    E[cnp].to = u, E[cnp].last = head[v], E[cnp].f = 0, E[cnp].c = -c; head[v] = cnp ++;
}

int Get_id(int x, int y)
{
    return (x - 1) * m + y;
}

void init()
{
    memset(head, -1, sizeof(head));
}

int SPFA()
{
    q.push_back(s);
    flow[s] = INF;
    for(int i = 1; i <= n * m * 3 + 3; i ++) dis[i] = INF; 
    while(!q.empty())
    {
        int u = q.front();
        q.pop_front();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = E[i].last)
        {
            int v = E[i].to;
            if(E[i].f && dis[v] > dis[u] + E[i].c)
            {
                dis[v] = dis[u] + E[i].c, pre[v] = i;
                flow[v] = min(flow[u], E[i].f);
                if(!vis[v])
                {
                    vis[v] = true;
                    if(!q.empty() && dis[v] < dis[q.front()]) q.push_front(v);
                    else q.push_front(v);
                }
            }
        }
    }
    if(dis[t] >= INF) return false;
    else return true;
}

void Max_flow()
{
    while(SPFA())
    {
        int v = pre[t];
        while(2333)
        {
            E[v].f -= flow[t];
            E[v ^ 1].f += flow[t];
            if(E[v ^ 1].to == s) break;
            v = pre[E[v ^ 1].to];
        }
        ans += flow[t];
        cost += flow[t] * dis[t];
    }
}

void Get_input()
{
    for(int i = 1; i <= n; i ++)
    {
        string s; cin >> s;
        for(int j = 0; j < m; j ++)
            a[i][j + 1] = s[j] - '0';
    }
    for(int i = 1; i <= n; i ++)
    {
        string s; cin >> s;
        for(int j = 0; j < m; j ++)
            b[i][j + 1] = s[j] - '0';
    }
    for(int i = 1; i <= n; i ++)
    {
        string s; cin >> s;
        for(int j = 0; j < m; j ++)
            c[i][j + 1] = s[j] - '0';
    }
}

void Connect()
{
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
        {
            int u = Get_id(i, j);
            if(a[i][j]) tem ++, add(s, u, 1, 0);
            if(b[i][j]) fans ++, add(u, t, 1, 0);
            if(a[i][j] == b[i][j])
            {
                add(u + size, u, c[i][j] / 2, 1);
                add(u, u + 2 * size, c[i][j] / 2, 1);
            }
            else if(b[i][j])
            {
                add(u + size, u, (c[i][j] + 1) / 2, 1);
                add(u, u + 2 * size, c[i][j] / 2, 1);
            }
            else if(a[i][j])
            {
                add(u + size, u, c[i][j] / 2, 1);
                add(u, u + 2 * size, (c[i][j] + 1) / 2, 1);
            }
            for(int k = 1; k <= 8; k ++)
            {
                int x = i + dx[k], y = j + dy[k];
                if(x < 1 || x > n || y < 1 || y > m) continue;
                add(u + 2 * size, Get_id(x, y) + size, INF, 0);
            }
        }
}

int main()
{
    scanf("%d%d", &n, &m);
    init();
    t = n * m * 3 + 3, size = n * m;
    Get_input();
    Connect(); 
    if(tem != fans) 
    {
        printf("-1
");
        return 0;
    }
    Max_flow();
    if(ans == fans) printf("%d", cost >> 1);
    else printf("-1
");
    return 0;
}

【题解】CQOI2012交换棋子

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