【上下界网络流 二分】bzoj2406: 矩阵

感觉考试碰到上下界网络流也还是写不来啊

Description

Input

第一行两个数n、m，表示矩阵的大小。

接下来n行，每行m列，描述矩阵A。

最后一行两个数L，R。

Output

第一行，输出最小的答案；

HINT

对于100%的数据满足N,M<=200,0<=L<=R<=1000,0<=Aij<=1000

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题目分析

首先二分行列之差的最大值。

这一类行列上的问题，属于经典的网络流模型。将行列各自看成点，由S向这些点连容量为$[ΣA_{i,j}-x,ΣA_{i,j}+x]$的边，再在行列之间互相连$[L,R]$的边，那么最终每个点的流量就是其行/列的权值和。

于是问题就变成了判定有源汇上下界可行流。

 #include<bits/stdc++.h>
 const int maxn = ;
 const int maxm = ;
 const int INF = 2e9;

 struct Edge
 {
     int u,v,f,c;
     Edge(int a=, int b=, int c=, int d=):u(a),v(b),f(c),c(d) {}
 }edges[maxm];
 int edgeTot,head[maxn],nxt[maxm],lv[maxn],cur[maxn];
 int r[maxn],c[maxn],a[maxn][maxn];
 int n,m,S,T,SS,TT,lLim,rLim,ans;

 int read()
 {
     char ch = getchar();
     int num = , fl = ;
     for (; !isdigit(ch); ch=getchar())
         if (ch=='-') fl = -;
     for (; isdigit(ch); ch=getchar())
         num = (num<<)+(num<<)+ch-;
     return num*fl;
 }
 void addedge(int u, int v, int c)
 {
     edges[edgeTot] = Edge(u, v, , c), nxt[edgeTot] = head[u], head[u] = edgeTot++;
     edges[edgeTot] = Edge(v, u, , ), nxt[edgeTot] = head[v], head[v] = edgeTot++;
 }
 bool buildLevel()
 {
     std::queue<int> q;
     memset(lv, , sizeof lv);
     lv[S] = , q.push(S);
     for (int i=; i<=TT; i++) cur[i] = head[i];
     for (int tmp; q.size(); )
     {
         tmp = q.front(), q.pop();
         for (int i=head[tmp]; i!=-; i=nxt[i])
         {
             int v = edges[i].v;
             if (!lv[v]&&edges[i].f < edges[i].c){
                 lv[v] = lv[tmp]+, q.push(v);
                 if (v==T) return true;
             }
         }
     }
     return false;
 }
 int fndPath(int x, int lim)
 {
     if (x==T) return lim;
     for (int &i=cur[x]; i!=-; i=nxt[i])
     {
         int v = edges[i].v, val;
         if (lv[x]+==lv[v]&&edges[i].f < edges[i].c){
             if ((val = fndPath(v, std::min(lim, edges[i].c-edges[i].f)))){
                 edges[i].f += val, edges[i^].f -= val;
                 return val;
             }else lv[v] = -;
         }
     }
     cur[x] = head[x];
     return ;
 }
 int dinic()
 {
     int ret = , val;
     while (buildLevel())
         while ((val = fndPath(S, INF))) ret += val;
     return ret;
 }
 bool check(int x)
 {
     int cur = ;
     memset(head, -, sizeof head);
     edgeTot = ;
     for (int i=; i<=n; i++)
     {
         if (r[i]+x < ) return false;
         if (r[i]-x > ){
             addedge(SS, T, r[i]-x);
             addedge(S, i, r[i]-x);
             addedge(SS, i, x<<);
             cur += r[i]-x;
         }else addedge(SS, i, r[i]+x);
     }
     for (int i=; i<=m; i++)
     {
         if (c[i]+x < ) return false;
         if (c[i]-x > ){
             addedge(S, TT, c[i]-x);
             addedge(i+n, T, c[i]-x);
             addedge(i+n, TT, x<<);
             cur += c[i]-x;
         }else addedge(i+n, TT, c[i]+x);
     }
     for (int i=; i<=n; i++)
         for (int j=; j<=m; j++)
             addedge(i, j+n, rLim);
     addedge(TT, SS, INF);
     return cur==dinic();
 }
 int main()
 {
     n = read(), m = read();
     S = n+m+, T = S+, SS = T+, TT = SS+;
     for (int i=; i<=n; i++)
         for (int j=; j<=m; j++)
             a[i][j] = read();
     lLim = read(), rLim = read();
     for (int i=; i<=n; i++)
         for (int j=; j<=m; j++)
             a[i][j] -= lLim, r[i] += a[i][j], c[j] += a[i][j];
     rLim -= lLim;
     int L = , R = std::max(*std::max_element(r+, r+n+), *std::max_element(c+, c+m+));
     for (int mid=(L+R)>>; L<=R; mid=(L+R)>>)
         if (check(mid)) ans = mid, R = mid-;
         else L = mid+;
     printf("%d\n",ans);
     return ;
 }

END