hdu 1569 &1565 (二分图带权最大独立集 - 最小割应用)

Posted 2021-01-07 xiuwenli

要选出一些点,这些点之间没有相邻边且要求权值之和最大,求这个权值
分析:二分图带权最大独立集.
用最大流最小割定理求解.其建图思路是:将所有格点编号,奇数视作X部,偶数视作Y部,建立源点S和汇点T, S向X部的点建边,Y部向T建边,容量为该点权值.
相邻的一对点(肯定是一奇一偶),由X中的点向Y中的点建边,容量为正无穷.
最后跑出最大流,|带权最大独立集| = |点权之和| - |最小割| = |点权之和| - |最大流|

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<string>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN=3010;//点数的最大值
const int MAXM=400010;//边数的最大值
#define captype int

struct SAP_MaxFlow{
    struct Edge{
        int from,to,next;
        captype cap;
    }edges[MAXM];
    int tot,head[MAXN];
    int gap[MAXN];
    int dis[MAXN];
    int cur[MAXN];
    int pre[MAXN];

    void init(){
        tot=0;
        memset(head,-1,sizeof(head));
    }
    void AddEdge(int u,int v,captype c,captype rc=0){
        edges[tot] = (Edge){u,v,head[u],c};  head[u]=tot++;
        edges[tot] = (Edge){v,u,head[v],rc}; head[v]=tot++;
    }
    captype maxFlow_sap(int sNode,int eNode, int n){//n是包括源点和汇点的总点个数，这个一定要注意
        memset(gap,0,sizeof(gap));
        memset(dis,0,sizeof(dis));
        memcpy(cur,head,sizeof(head));
        pre[sNode] = -1;
        gap[0]=n;
        captype ans=0;
        int u=sNode;
        while(dis[sNode]<n){
            if(u==eNode){
                captype Min=INF ;
                int inser;
                for(int i=pre[u]; i!=-1; i=pre[edges[i^1].to])
                if(Min>edges[i].cap){
                    Min=edges[i].cap;
                    inser=i;
                }
                for(int i=pre[u]; i!=-1; i=pre[edges[i^1].to]){
                    edges[i].cap-=Min;
                    edges[i^1].cap+=Min;
                }
                ans+=Min;
                u=edges[inser^1].to;
                continue;
            }
            bool flag = false;
            int v;
            for(int i=cur[u]; i!=-1; i=edges[i].next){
                v=edges[i].to;
                if(edges[i].cap>0 && dis[u]==dis[v]+1){
                    flag=true;
                    cur[u]=pre[v]=i;
                    break;
                }
            }
            if(flag){
                u=v;
                continue;
            }
            int Mind= n;
            for(int i=head[u]; i!=-1; i=edges[i].next)
            if(edges[i].cap>0 && Mind>dis[edges[i].to]){
                Mind=dis[edges[i].to];
                cur[u]=i;
            }
            gap[dis[u]]--;
            if(gap[dis[u]]==0) return ans;
            dis[u]=Mind+1;
            gap[dis[u]]++;
            if(u!=sNode) u=edges[pre[u]^1].to;  //退一条边
        }
        return ans;
    }
}F;

int G[55][55];
int dir[4][2] = {1,0,-1,0,0,1,0,-1};


int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int N,M;
    int u,v,tmp;
    while(scanf("%d %d",&N, &M)==2){
        int S = 0,T = N*M+1;
        int sum  =0;
        F.init();
        for(int i=1;i<=N;++i){
            for(int j=1;j<=M;++j){
                scanf("%d",&tmp);
                G[i][j] = tmp;
                sum+=tmp;
            }
        }
        for(int i=1;i<=N;++i){
            for(int j=1;j<=M;++j){
                int id = (i-1)*M +j;
                if((i+j)&1){
                    F.AddEdge(S,id,G[i][j]);
                    for(int k=0;k<4;++k){
                        int nx = i+ dir[k][0];
                        int ny = j+ dir[k][1];
                        if(nx<1 || nx>N || ny<1 ||ny>M) continue;
                        int nid = (nx-1)*M + ny;
                        F.AddEdge(id,nid,INF);
                    }
                }
                else{
                    F.AddEdge((i-1)*M+j,T,G[i][j]);
                }
            }
        }
        int flow = F.maxFlow_sap(S,T,T+1);
        int res = sum - flow;
        printf("%d
",res);
    }
    return 0;
}