P4313 文理分科

思路

遇到这种利益冲突的最终利益最大化问题
考虑转化为最小割，使得损失的价值最小
相当于文科是S，理科是T，选出最小割就是确定损失代价最小的方案
然后就把S向每个点连一条cap=art[i][j]的边，每个点向T连一条cap=science[i][j]的边，再新建n*m个点表示同选文科的利益，然后S向每个新建点连一条cap=same_art[i][j]的边，然后再从新建点向每个点和它的相邻点向连一条cap=INF的边，然后同选立刻的收益同理新建点再向T连边，相邻点同理的向新建点连边
再跑出最小割即可

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
struct Edge{
    int u,v,cap,flow;
};
const int MAXN = 40100;
const int INF = 0x3f3f3f3f;
vector<Edge> edges;
vector<int> G[MAXN];
void addedge(int u,int v,int cap){
    edges.push_back((Edge){u,v,cap,0});
    edges.push_back((Edge){v,u,0,0});
    int cnt=edges.size();
    G[u].push_back(cnt-2);
    G[v].push_back(cnt-1);
}
int cur[MAXN],dep[MAXN],vis[MAXN],s,t;
int dfs(int x,int a){
    if(x==t||a==0)
        return a;
    int f,flow=0;
    for(int &i=cur[x];i<G[x].size();i++){
        Edge &e = edges[G[x][i]];
        if(dep[e.v]==dep[x]+1&&(f=dfs(e.v,min(e.cap-e.flow,a)))>0){
            flow+=f;
            e.flow+=f;
            edges[G[x][i]^1].flow-=f;
            a-=f;
            if(!a)
                break;
        }
    }
    return flow;
}
queue<int> q;
bool bfs(void){
    memset(vis,0,sizeof(vis));
    dep[s]=0;
    vis[s]=true;
    q.push(s);
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=0;i<G[x].size();i++){
            Edge &e = edges[G[x][i]];
            if(e.cap>e.flow&&(!vis[e.v])){
                vis[e.v]=true;
                dep[e.v]=dep[x]+1;
                q.push(e.v);
            }
        }
    }
    return vis[t];
}
int dinic(void){
    int flow=0;
    while(bfs()){
            // printf("Not Re
");
        memset(cur,0,sizeof(cur));
        flow+=dfs(s,INF);
    }
    return flow;
}
int n,m,same_wen[110][110],same_li[110][110],wen[110][110],li[110][110],sum=0;
const int mx[] = {0,0,1,-1,0},my[] ={0,1,0,0,-1};
int id(int x,int y,int idx=0){
    return (x-1)*m+y+idx*n*m;
}
int main(){
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            scanf("%d",&wen[i][j]);
            sum+=wen[i][j];
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            scanf("%d",&li[i][j]);
            sum+=li[i][j];
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            scanf("%d",&same_wen[i][j]);
            sum+=same_wen[i][j];
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            scanf("%d",&same_li[i][j]);
            sum+=same_li[i][j];
        }
    // printf("Not Re
");
    s=MAXN-2;//wen
    t=MAXN-3;//li
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            addedge(s,id(i,j),wen[i][j]);//wen
            addedge(id(i,j),t,li[i][j]);//li
        }
    for(int i=1;i<=n;i++)//shang
        for(int j=1;j<=m;j++){
            addedge(s,id(i,j,1),same_wen[i][j]);
            addedge(id(i,j,1),id(i,j),INF);
            if(i>1)
                addedge(id(i,j,1),id(i-1,j),INF);
            if(j>1)
                addedge(id(i,j,1),id(i,j-1),INF);
            if(i<n)
                addedge(id(i,j,1),id(i+1,j),INF);
            if(j<m)
                addedge(id(i,j,1),id(i,j+1),INF);
        }
    for(int i=1;i<=n;i++)//shang
        for(int j=1;j<=m;j++){
            addedge(id(i,j,2),t,same_li[i][j]);
            addedge(id(i,j),id(i,j,2),INF);
            if(i>1)
                addedge(id(i-1,j),id(i,j,2),INF);
            if(j>1)
                addedge(id(i,j-1),id(i,j,2),INF);
            if(i<n)
                addedge(id(i+1,j),id(i,j,2),INF);
            if(j<m)
                addedge(id(i,j+1),id(i,j,2),INF);
        }
    // printf("Not Re
");
    printf("%d
",sum-dinic());
    return 0;
}