BZOJ 2406: 矩阵 [上下界网络流 二分答案]

2406: 矩阵

题意：自己去看吧,最小化每行每列所有元素与给定矩阵差的和的绝对值中的最大值

又带绝对值又带max不方便直接求

显然可以二分这个最大值

然后判定问题，给定矩阵每行每列的范围和每个元素的取值范围判断可行...和之前做过的一样了上下界可行流

1A好开心啊

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define fir first
#define sec second
typedef long long ll;
const int N=2005, M=1e5+5, INF=1e9;
inline ll read(){
char c=getchar();ll x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}

int n, m, l, r, a, row
, col
, s, t, extra
, tot;

struct edge{int v, c, f, ne;}e[M];
int cnt=1, h
;
inline void ins(int u, int v, int c) {
e[++cnt]=(edge){v, c, 0, h[u]}; h[u]=cnt;
e[++cnt]=(edge){u, 0, 0, h[v]}; h[v]=cnt;
}

int q
, head, tail, vis
, d
, cur
;
bool bfs(int s, int t) {
memset(vis, 0, sizeof(vis));
head=tail=1;
q[tail++]=s; d[s]=0; vis[s]=1;
while(head!=tail) {
int u=q[head++];
for(int i=h[u];i;i=e[i].ne)
if(!vis[e[i].v] && e[i].c>e[i].f) {
vis[e[i].v]=1; d[e[i].v]=d[u]+1;
q[tail++]=e[i].v;
if(e[i].v == t) return true;
}
}
return false;
}
int dfs(int u, int a, int t) {
if(u==t || a==0) return a;
int flow=0, f;
for(int &i=cur[u];i;i=e[i].ne)
if(d[e[i].v]==d[u]+1 && (f=dfs(e[i].v, min(a, e[i].c-e[i].f), t))>0) {
flow+=f;
e[i].f+=f;
e[i^1].f-=f;
a-=f;
if(a==0) break;
}
if(a) d[u]=-1;
return flow;
}
int dinic(int s, int t) {
int flow=0;
while(bfs(s, t)) {
for(int i=0; i<=tot; i++) cur[i]=h[i];
flow+=dfs(s, INF, t);
}
return flow;
}

bool check(int mid) { //printf("\ncheck %d\n",mid);
s=0; t=n+m+1;
cnt=1; memset(h,0,sizeof(h)); memset(extra,0,sizeof(extra));
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++) ins(i, n+j, r-l), extra[i]-=l, extra[n+j]+=l;
for(int i=1; i<=n; i++) {
int l = -mid+row[i], r = mid+row[i];
ins(s, i, r-l), extra[s]-=l, extra[i]+=l;
}
for(int j=1; j<=m; j++) {
int l = -mid+col[j], r = mid+col[j];
ins(n+j, t, r-l), extra[n+j]-=l, extra[t]+=l;
}
int ss=t+1, tt=t+2, sum=0; tot=tt;
for(int i=s; i<=t; i++) {
if(extra[i]>0) ins(ss, i, extra[i]), sum+=extra[i];
if(extra[i]<0) ins(i, tt, -extra[i]);
}
ins(t, s, INF);

int flow=dinic(ss, tt); //printf("flow %d %d\n",flow,sum);
return flow==sum;
}
void solve() {
int l=0, r=200000, ans=-1;
while(l<=r) {
int mid=(l+r)>>1;
if(check(mid)) ans=mid, r=mid-1;
else l=mid+1;
}
printf("%d",ans);
}
int main() {
freopen("in","r",stdin);
n=read(); m=read();
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++) a=read(), row[i]+=a, col[j]+=a;

l=read(); r=read();
solve();
}