bzoj 2406: 矩阵 （有上下界的网络流）

2406: 矩阵

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 229  Solved: 90
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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；

Sample Input

2 2

0 1

2 1

0 1

Sample Output

1

HINT

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

Source

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题解：有上下界的网络流

直接二分要求的最小答案x，然后用有源汇有上下界的可行流判断。

s->i 容量[sumi[i]-x,sumi[i]+x] sumi[i]表示第i行的总和

j->t 容量[sumj[j]-x,sumj[j]+x] sumj[j]表示第j行的总和

i->j 容量[l,r]

然后按照有源汇有上下界的可行流建模即可。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#include<queue>#define N 200000using namespace std;int in[N],out[N],point[N],v[N],remain[N],nxt[N];int deep[N],num[N],cur[N],ls,rs,n,m,tot;int a[203][203],sumi[N],sumj[N],last[N],ck[N];void add(int x,int y,int z){if (!z) return;tot++; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=z;tot++; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;//cout<<x<<" "<<y<<" "<<z<<endl;}int addflow(int s,int t){int ans=1e9; int now=t;while (now!=s) {ans=min(ans,remain[last[now]]);now=v[last[now]^1];}now=t;while (now!=s){remain[last[now]]-=ans;remain[last[now]^1]+=ans;now=v[last[now]^1];}return ans;}void bfs(int s,int t){for (int i=1;i<=t;i++) deep[i]=t;deep[t]=0; queue<int> p; p.push(t);while (!p.empty()) {int now=p.front(); p.pop();for (int i=point[now];i!=-1;i=nxt[i]) if (deep[v[i]]==t&&remain[i^1])  deep[v[i]]=deep[now]+1,p.push(v[i]);}}void isap(int s,int t){bfs(s,t);int ans=0,now=s;for (int i=1;i<=t;i++) cur[i]=point[i];for (int i=1;i<=t;i++) num[deep[i]]++;while (deep[s]<t) {if (now==t) {ans+=addflow(s,t);    now=s;}bool pd=false;for (int i=cur[now];i!=-1;i=nxt[i])  if (deep[v[i]]+1==deep[now]&&remain[i]) { pd=true; cur[now]=i; last[v[i]]=i; now=v[i]; break; }if (!pd) {int minn=t;    for (int i=point[now];i!=-1;i=nxt[i])      if (remain[i]) minn=min(minn,deep[v[i]]);    if (!--num[deep[now]]) break;    num[deep[now]=minn+1]++;    cur[now]=point[now];    if (now!=s) now=v[last[now]^1];}}}bool check(){for (int i=1;i<=n+m+2;i++) if (remain[ck[i]^1]) return false;return true;}bool solve(int x){tot=-1;memset(point,-1,sizeof(point));memset(num,0,sizeof(num));memset(in,0,sizeof(in));memset(out,0,sizeof(out));int s=1; int t=n+m+2;for (int i=1;i<=n;i++) {add(s,i+1,2*x);in[i+1]+=sumi[i]-x;out[s]+=sumi[i]-x;}for (int i=1;i<=m;i++) {add(i+n+1,t,2*x);in[t]+=sumj[i]-x;out[i+n+1]+=sumj[i]-x;}for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) { add(i+1,j+n+1,rs-ls); in[j+n+1]+=ls; out[i+1]+=ls; }int ss=t+1; int tt=ss+1;for (int i=1;i<=t;i++){int k=out[i]-in[i];if (k>=0) add(i,tt,k);else add(ss,i,-k);ck[i]=tot;}add(t,s,1e9);isap(ss,tt);//for (int i=1;i<=n+m+1;i++) //cout<<remain[ck[i]^1]<<" ";//cout<<endl;if (check()) return true;return false;}int main(){freopen("a.in","r",stdin);scanf("%d%d",&n,&m);int sum=0;for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) {  scanf("%d",&a[i][j]);  sum+=a[i][j];  sumi[i]+=a[i][j]; sumj[j]+=a[i][j];    }scanf("%d%d",&ls,&rs);int l=0; int r=sum; int ans;//cout<<ls<<" "<<rs<<endl;for (int i=1;i<=n;i++) r=min(r,sumi[i]);for (int i=1;i<=m;i++) r=min(r,sumj[i]);ans=r;//cout<<l<<" "<<r<<endl;while (l<=r) {int mid=(l+r)/2;if (solve(mid)) ans=min(ans,mid),r=mid-1;else l=mid+1; }printf("%d\n",ans);}