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=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;
    for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
    if (ok) x=-x;
}
const int maxn=2005;
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-1]-sum[x-1][yy]+sum[x-1][y-1];}
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*2){
        if (x==xx) y++;
        else if (calc(x+1,y,xx,yy)>=k) x++;
        else xx--;
    }
    printf("%d %d %d %d
",y,x,yy,xx);
    exit(0);
}
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=0;
    for (int i=1;i<=n+1;i++){
        while (top&&stack[top].val>h[i]) r[stack[top--].id]=i-1;
        stack[++top]=(Data){h[i],i};
    }
    top=0;
    for (int i=n;i>=0;i--){
        while (top&&stack[top].val>h[i]) l[stack[top--].id]=i+1;
        stack[++top]=(Data){h[i],i};
    }
    for (int i=1;i<=n;i++)
        if (h[i]) if (calc(x-h[i]+1,l[i],x,r[i])>=k) output(x-h[i]+1,l[i],x,r[i]);
}
int main(){
    read(k),read(n);
    for (int i=1;i<=n;i++) for (int j=1;j<=n;j++){
        read(a[i][j]),sum[i][j]=sum[i][j-1]+a[i][j];
        if (k<=a[i][j]&&a[i][j]<=k*2){printf("%d %d %d %d
",j,i,j,i);return 0;}
    }
    for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) sum[i][j]+=sum[i-1][j];
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++) h[j]=a[i][j]>k*2?0:h[j]+1;
        work(i);
    }
    puts("NIE");
    return 0;
}