BZOJ2756: [SCOI2012]奇怪的游戏

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2756

对棋盘进行黑白染色。若答案为x，有c1*x-s1==c2*x-s2得x=(s1-s2)/(c1-c2)，那么若c1!=c2,检测x就可以了。若不是就二分x。

建图：s->黑点 c=x-a[i][j] 白点->t c=x-a[i][j] 黑->白 c=inf

#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 2005
#define eps 1e-3
#define ll long long
#define inf (1LL<<50)
using namespace std;
struct data{int obj,pre;ll c;
}e[200500];
int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
int head[maxn],cur[maxn],uu[maxn],n,m,t,tot,s;
ll c1,c2,s1,s2,a[42][42];
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch==‘-‘) f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-‘0‘; ch=getchar();}
    return x*f;
}
void insert(int x,int y,ll z){
    e[++tot].obj=y; e[tot].c=z; e[tot].pre=head[x]; head[x]=tot;
    e[++tot].obj=x; e[tot].c=0; e[tot].pre=head[y]; head[y]=tot;
}
int p(int x,int y){
    return (x-1)*m+y;
}
bool bfs(){
    queue<int> q; q.push(0); clr(uu,-1); uu[0]=0;
    while (!q.empty()){
        int u=q.front(); q.pop();
        for (int j=head[u];j;j=e[j].pre){
            int v=e[j].obj;
            if (uu[v]==-1&&e[j].c) {
                uu[v]=uu[u]+1; q.push(v);
            }
        }
    }
    if (uu[t]==-1) return 0;
    return 1;
}
ll dfs(int x,ll mx){
    if (x==t) return mx;
    ll used=0;
    for (int j=cur[x];j;j=e[j].pre){
        int v=e[j].obj;
        if (uu[v]==uu[x]+1){
            ll w=dfs(v,min(e[j].c,mx-used));
            e[j].c-=w; e[j^1].c+=w; used+=w;
            if (e[j].c) cur[x]=j;
            if (used==mx) return mx;
        }
    }
    if (!used) uu[x]=-1;
    return used;
}
ll dinic(){
    ll ans=0;
    while (bfs()){
        rep(i,0,t) cur[i]=head[i];
        ans+=dfs(0,inf);
    }
    return ans;
}
bool jud(ll x){
    tot=1; clr(head,0); s=0; t=n*m+1;
    ll cnt=0;
    rep(i,1,n) rep(j,1,m) {
        if ((i+j)&1) {
            insert(0,p(i,j),x-a[i][j]),cnt+=x-a[i][j];
            rep(k,0,3) {
                 int x=i+dx[k],y=j+dy[k];
                if (x<1||x>n||y<1||y>m) continue;
                insert(p(i,j),p(x,y),inf);
             }
        }
        else insert(p(i,j),t,x-a[i][j]);
     }
     if (dinic()!=cnt) return 0;
    return 1;
}
int main(){
    int T=read();
    while (T--){
        ll mx=0; c1=0,c2=0,s1=0,s2=0;
        n=read(); m=read();
        rep(i,1,n) rep(j,1,m){
            a[i][j]=read(); mx=max(mx,a[i][j]);
            if ((i+j)&1) c1++,s1+=a[i][j];
            else c2++,s2+=a[i][j];
        }
        if (c1!=c2){
            ll x=(s1-s2)/(c1-c2);
            if (x>=mx) if (jud(x)) {printf("%lld\n",x*c1-s1); continue;}
            puts("-1");
        }
        else {
            ll l=mx,r=inf;
            while (l<=r){
                ll mid=(l+r)/2;
                if (jud(mid)) r=mid-1;
                else l=mid+1;
            }
            printf("%lld\n",l*c1-s1);
        }
    }
    return 0;
}