「网络流 24 题」方格取数

大意: 给定$n*m$棋盘, 每个格子有权值, 不能选择相邻格子, 求能选出的最大权值.

二分图带权最大独立集, 转化为最小割问题.

S与$X$连边权为权值的边, $X$与$Y$之间连$INF$, $Y$与$T$连边权为权值的边.

则最大权值为总权值-最小割. 残量网络中与$S$相连的或与$T$相连的表示选择, 否则表示不选.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#include <unordered_map>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head

const int N = 1e6+10;
const int dx[]={0,0,1,-1};
const int dy[]={1,-1,0,0};
int n, m, S, T;
struct edge {
    int v,w,next;
} e[N];
int head[N], dep[N], vis[N], cur[N], cnt=1;
queue<int> Q;
void add(int u, int v, int w) {
    e[++cnt] = {v,w,head[u]};
    head[u] = cnt;
    e[++cnt] = {u,0,head[v]};
    head[v] = cnt;
}
int bfs() {
    REP(i,1,T) dep[i]=INF,vis[i]=0,cur[i]=head[i];
    dep[S]=0,Q.push(S);
    while (Q.size()) {
        int u = Q.front(); Q.pop();
        for (int i=head[u]; i; i=e[i].next) {
            if (dep[e[i].v]>dep[u]+1&&e[i].w) {
                dep[e[i].v]=dep[u]+1;
                Q.push(e[i].v);
            }
        }
    }
    return dep[T]!=INF;
}
int dfs(int x, int w) {
    if (x==T) return w;
    int used = 0;
    for (int i=cur[x]; i; i=e[i].next) {
        cur[x] = i;
        if (dep[e[i].v]==dep[x]+1&&e[i].w) {
            int flow = dfs(e[i].v,min(w-used,e[i].w));
            if (flow) {
                used += flow;
                e[i].w -= flow;
                e[i^1].w += flow;
                if (used==w) break;
            }
        }
    }
    return used;
}

int dinic() {
    int ans = 0;
    while (bfs()) ans+=dfs(S,INF);
    return ans;
}

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

int main() {
	scanf("%d%d", &n, &m);
	S = n*m+1, T = S+1;
	int tot = 0;
	REP(i,1,n) REP(j,1,m) {
		int t;
		scanf("%d", &t);
		if (i+j&1) {
			add(S,get(i,j),t);
			REP(k,0,3) {
				int ii=i+dx[k],jj=j+dy[k];
				if (1<=ii&&ii<=n&&1<=jj&&jj<=m) {
					add(get(i,j),get(ii,jj),INF);
				}
			}
		}
		else add(get(i,j),T,t);
		tot += t;
	}
	printf("%d\n",tot-dinic());
}

原文地址：https://www.cnblogs.com/uid001/p/10987334.html