SCU3033 Destroying a Painting（最小费用最大流）

题目大概说有一个有n*m个格子的画板，画板上每个格子都有颜色，现在要把所有格子的颜色改成红、绿或者蓝，改变的代价是二者RGB值的曼哈顿距离，还要求红绿蓝格子个数的最大值和最小值要尽可能接近，问最少的代价是多少。

红绿蓝三色的个数是可以直接确定的，分别考虑几个情况就OK了，然后就是根据红绿蓝的个数构图跑最小费用最大流。。

 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<algorithm>
 using namespace std;
 #define INF (1<<30)
 #define MAXN 444
 #define MAXM 444*888
 struct Edge{
     int u,v,cap,cost,next;
 }edge[MAXM];
 int head[MAXN];
 int NV,NE,vs,vt;

 void addEdge(int u,int v,int cap,int cost){
     edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost;
     edge[NE].next=head[u]; head[u]=NE++;
     edge[NE].u=v; edge[NE].v=u; edge[NE].cap=; edge[NE].cost=-cost;
     edge[NE].next=head[v]; head[v]=NE++;
 }
 bool vis[MAXN];
 int d[MAXN],pre[MAXN];
 bool SPFA(){
     for(int i=;i<NV;++i){
         vis[i]=;
         d[i]=INF;
     }
     vis[vs]=;
     d[vs]=;
     queue<int> que;
     que.push(vs);
     while(!que.empty()){
         int u=que.front(); que.pop();
         for(int i=head[u]; i!=-; i=edge[i].next){
             int v=edge[i].v;
             if(edge[i].cap && d[v]>d[u]+edge[i].cost){
                 d[v]=d[u]+edge[i].cost;
                 pre[v]=i;
                 if(!vis[v]){
                     vis[v]=;
                     que.push(v);
                 }
             }
         }
         vis[u]=;
     }
     return d[vt]!=INF;
 }
 int MCMF(){
     int res=;
     while(SPFA()){
         int flow=INF,cost=;
         for(int u=vt; u!=vs; u=edge[pre[u]].u){
             flow=min(flow,edge[pre[u]].cap);
         }
         for(int u=vt; u!=vs; u=edge[pre[u]].u){
             edge[pre[u]].cap-=flow;
             edge[pre[u]^].cap+=flow;
             cost+=flow*edge[pre[u]].cost;
         }
         res+=cost;
     }
     return res;
 }

 inline void in(int &ret){
     char c; ret=;
     while(c=getchar(),c<''||c>'');
     while(c>=''&&c<='') ret=ret*+(c-''),c=getchar();
 }
 int n,m,R[][],G[][],B[][];
 int get(int r,int g,int b){
     vs=n*m+; vt=vs+; NV=vt+; NE=;
     memset(head,-,sizeof(head));
     addEdge(n*m,vt,r,);
     addEdge(n*m+,vt,g,);
     addEdge(n*m+,vt,b,);
     for(int i=; i<n; ++i){
         for(int j=; j<m; ++j){
             addEdge(vs,i*m+j,,);
             addEdge(i*m+j,n*m,,abs(-R[i][j])+G[i][j]+B[i][j]);
             addEdge(i*m+j,n*m+,,R[i][j]+abs(-G[i][j])+B[i][j]);
             addEdge(i*m+j,n*m+,,R[i][j]+G[i][j]+abs(-B[i][j]));
         }
     }
     return MCMF();
 }
 int main(){
     int t;
     in(t);
     for(int cse=; cse<=t; ++cse){
         in(n); in(m);
         for(int i=; i<n; ++i){
             for(int j=; j<m; ++j){
                 in(R[i][j]); in(G[i][j]); in(B[i][j]);
             }
         }
         int ans=INF;
         if(n*m%==){
             ans=min(ans,get(n*m/,n*m/,n*m/));
         }else if(n*m%==){
             ans=min(ans,get(n*m/+,n*m/,n*m/));
             ans=min(ans,get(n*m/,n*m/+,n*m/));
             ans=min(ans,get(n*m/,n*m/,n*m/+));
         }else{
             ans=min(ans,get(n*m/+,n*m/+,n*m/));
             ans=min(ans,get(n*m/+,n*m/,n*m/+));
             ans=min(ans,get(n*m/,n*m/+,n*m/+));
         }
         printf("Case %d: %d\n",cse,ans);
     }
     return ;
 }