• 方格取数问题


    又是套路啊
    黑白染色,S,T连不同色的,求最小割,用总和-最小割即为答案

    # include <stdio.h>
    # include <stdlib.h>
    # include <iostream>
    # include <string.h>
    # include <math.h>
    # define ll long long
    # define RG register
    # define IL inline
    # define Mem(a, b) memset(a, b, sizeof(a))
    # define Max(a, b) (((a) > (b)) ? (a) : (b))
    # define Min(a, b) (((a) < (b)) ? (a) : (b))
    using namespace std;
    
    IL int Get(){
        RG char c = '!'; RG int x = 0, z;
        for(; c > '9' || c < '0'; c = getchar()) z = (c == '-') ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
        return x * z;
    }
    
    const int INF = 2147483647, MAXN = 50001, MAXM = 2000001;
    int n, m, ft[MAXN], cnt, ans, map[31][31], Q[MAXM], level[MAXN], cur[MAXN];
    int dx[] = {-1, 0, 1, 0};
    int dy[] = {0, 1, 0, -1};
    struct Edge{
        int to, nt, f;
    } edge[MAXM];
    
    IL void _Add(RG int u, RG int v, RG int f){
        edge[cnt] = (Edge){v, ft[u], f}; ft[u] = cnt++;
        edge[cnt] = (Edge){u, ft[v], 0}; ft[v] = cnt++;
    }
    
    IL void Add(RG int x, RG int y){
        RG int u = (x - 1) * m + y;
        for(RG int i = 0; i < 4; i++){
            RG int xx = x + dx[i], yy = y + dy[i];
            if(xx <= 0 || xx > n || yy <= 0 || yy > m) continue;
            RG int v = (xx - 1) * m + yy;
            _Add(u, v, INF);
        }
    }
    
    IL int Dfs(RG int u, RG int T, RG int maxf){
        if(u == T) return maxf;
        RG int ret = 0;
        for(RG int &e = cur[u]; e != -1; e = edge[e].nt){
            RG int v = edge[e].to;
            if(level[v] == level[u] + 1 && edge[e].f){
                RG int f = Dfs(v, T, Min(maxf - ret, edge[e].f));
                edge[e].f -= f; edge[e ^ 1].f += f;
                ret += f;
                if(ret == maxf) break;
            }
        }
        return ret;
    }
    
    IL bool Bfs(RG int S, RG int T){
        Mem(level, 0);
        RG int head = 0, tail = 0;
        Q[0] = S; level[S] = 1;
        while(head <= tail){
            RG int u = Q[head++];
            if(u == T) return 1;
            for(RG int e = ft[u]; e != -1; e = edge[e].nt){
                RG int v = edge[e].to;
                if(!level[v] && edge[e].f){
                    level[v] = level[u] + 1;
                    Q[++tail] = v;
                }
            }
        }
        return 0;
    }
    
    int main(){
        Mem(ft, -1);
        n = Get(); m = Get();
        RG int S = 0, T = n * m + 1, tot = 0;
        for(RG int i = 1; i <= n; i++)
            for(RG int j = 1; j <= m; j++){
                map[i][j] = Get(); tot += map[i][j];
                if(~(i + j) & 1) _Add(S, (i - 1) * m + j, map[i][j]), Add(i, j);
                else _Add((i - 1) * m + j, T, map[i][j]);
            }
        while(Bfs(S, T)){
            for(RG int i = S; i <= T; i++)
                cur[i] = ft[i];
            ans += Dfs(S, T, INF);
        }
        printf("%d
    ", tot - ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8206329.html
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