【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

---

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 = 435;
const int maxm = 500035;
const int INF = 2e9;

struct Edge
{
    int u,v,f,c;
    Edge(int a=0, int b=0, int c=0, int d=0):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 = 0, fl = 1;
    for (; !isdigit(ch); ch=getchar())
        if (ch=='-') fl = -1;
    for (; isdigit(ch); ch=getchar())
        num = (num<<1)+(num<<3)+ch-48;
    return num*fl;
}
void addedge(int u, int v, int c)
{
    edges[edgeTot] = Edge(u, v, 0, c), nxt[edgeTot] = head[u], head[u] = edgeTot++;
    edges[edgeTot] = Edge(v, u, 0, 0), nxt[edgeTot] = head[v], head[v] = edgeTot++;
}
bool buildLevel()
{
    std::queue<int> q;
    memset(lv, 0, sizeof lv);
    lv[S] = 1, q.push(S);
    for (int i=1; i<=TT; i++) cur[i] = head[i];
    for (int tmp; q.size(); )
    {
        tmp = q.front(), q.pop();
        for (int i=head[tmp]; i!=-1; i=nxt[i])
        {
            int v = edges[i].v;
            if (!lv[v]&&edges[i].f < edges[i].c){
                lv[v] = lv[tmp]+1, 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!=-1; i=nxt[i])
    {
        int v = edges[i].v, val;
        if (lv[x]+1==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^1].f -= val;
                return val;
            }else lv[v] = -1;
        }
    }
    cur[x] = head[x];
    return 0;
}
int dinic()
{
    int ret = 0, val;
    while (buildLevel())
        while ((val = fndPath(S, INF))) ret += val;
    return ret;
}
bool check(int x)
{
    int cur = 0;
    memset(head, -1, sizeof head);
    edgeTot = 0;
    for (int i=1; i<=n; i++)
    {
        if (r[i]+x < 0) return false;
        if (r[i]-x > 0){
            addedge(SS, T, r[i]-x);
            addedge(S, i, r[i]-x);
            addedge(SS, i, x<<1);
            cur += r[i]-x;
        }else addedge(SS, i, r[i]+x);
    }
    for (int i=1; i<=m; i++)
    {
        if (c[i]+x < 0) return false;
        if (c[i]-x > 0){
            addedge(S, TT, c[i]-x);
            addedge(i+n, T, c[i]-x);
            addedge(i+n, TT, x<<1);
            cur += c[i]-x;
        }else addedge(i+n, TT, c[i]+x);
    }
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            addedge(i, j+n, rLim);
    addedge(TT, SS, INF);
    return cur==dinic();
}
int main()
{
    n = read(), m = read();
    S = n+m+1, T = S+1, SS = T+1, TT = SS+1;
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            a[i][j] = read();
    lLim = read(), rLim = read();
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            a[i][j] -= lLim, r[i] += a[i][j], c[j] += a[i][j];
    rLim -= lLim;
    int L = 0, R = std::max(*std::max_element(r+1, r+n+1), *std::max_element(c+1, c+m+1));
    for (int mid=(L+R)>>1; L<=R; mid=(L+R)>>1)
        if (check(mid)) ans = mid, R = mid-1;
        else L = mid+1;
    printf("%d
",ans);
    return 0;
}

 

 

 

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