bzoj1127: [POI2008]KUP

Description

给一个n*n的地图，每个格子有一个价格，找一个矩形区域，使其价格总和位于[k,2k]

Input

输入k n（n<2000)和一个n*n的地图

Output

输出矩形的左上和右下的列-行坐标或NIE

Sample Input

inputdata1
4 3
1 1 1
1 9 1
1 1 1
inputdata2
8 4
1 2 1 3
25 1 2 1
4 20 3 3
3 30 12 2

Sample Output

outputdata1
NIE
outputdata2
2 1 4 2

题解：

http://blog.csdn.net/popoqqq/article/details/44625423

code：

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 typedef long long int64;
 char ch;
 bool ok;
 void read(int &x){
     for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
     for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
     if (ok) x=-x;
 }
 const int maxn=;
 int k,n,a[maxn][maxn],head,tail,last,h[maxn],top,l[maxn],r[maxn];
 int64 sum[maxn][maxn];
 int64 calc(int x,int y,int xx,int yy){return sum[xx][yy]-sum[xx][y-]-sum[x-][yy]+sum[x-][y-];}
 struct Data{
     int val,id;
 }que[maxn],stack[maxn];
 void output(int x,int y,int xx,int yy){
     while (calc(x,y,xx,yy)>k*){
         if (x==xx) y++;
         else if (calc(x+,y,xx,yy)>=k) x++;
         else xx--;
     }
     printf("%d %d %d %d\n",y,x,yy,xx);
     exit();
 }
 void work(int x){
     /*head=1,tail=0,last=1;
     for (int i=1;i<=n;i++){
         while (head<=tail&&que[tail].val>=h[i]) tail--;
         que[++tail]=(Data){h[i],i};
         while (head<=tail&&que[head].val==0) last=que[head++].id+1;
         if (head<=tail&&calc(x-que[head].val+1,last,x,i)>=k) output(x-que[head].val+1,last,x,i);
     }*/
     top=;
     for (int i=;i<=n+;i++){
         while (top&&stack[top].val>h[i]) r[stack[top--].id]=i-;
         stack[++top]=(Data){h[i],i};
     }
     top=;
     for (int i=n;i>=;i--){
         while (top&&stack[top].val>h[i]) l[stack[top--].id]=i+;
         stack[++top]=(Data){h[i],i};
     }
     for (int i=;i<=n;i++)
         if (h[i]) if (calc(x-h[i]+,l[i],x,r[i])>=k) output(x-h[i]+,l[i],x,r[i]);
 }
 int main(){
     read(k),read(n);
     for (int i=;i<=n;i++) for (int j=;j<=n;j++){
         read(a[i][j]),sum[i][j]=sum[i][j-]+a[i][j];
         if (k<=a[i][j]&&a[i][j]<=k*){printf("%d %d %d %d\n",j,i,j,i);return ;}
     }
     for (int i=;i<=n;i++) for (int j=;j<=n;j++) sum[i][j]+=sum[i-][j];
     for (int i=;i<=n;i++){
         for (int j=;j<=n;j++) h[j]=a[i][j]>k*?:h[j]+;
         work(i);
     }
     puts("NIE");
     return ;
 }