您的位置:首页 > 其它

URAL 1097. Square Country 2 (离散化,最大正方形面积)

2013-08-24 10:50 260 查看
题意:http://www.nocow.cn/index.php/Translate:URAL/1097

离散化之后找最大没有被覆盖的正方形。暴力能过。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <string>

#define LL long long
#define DB double
#define SI(a) scanf("%d",&a)
#define SD(a) scanf("%lf",&a)
#define SS(a) scanf("%s",a)
#define SF scanf
#define PF printf
#define MM(a,v) memset(a,v,sizeof(a))
#define REP(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define REPD(i,a,b) for(int (i)=(a);(i)>(b);(i)--)
#define N 209
#define INF 0x3f3f3f3f
#define EPS 1e-8
#define bug puts("bug")
using namespace std;
int v

;
int x
,y
;
int n,m;
int k;
int lx=0,ly=0;
struct R{
    int val,x,y,l;
    void get(){
        SI(val);SI(l);SI(x);SI(y);
    }
} re
,tmp
;
int finx(int t)
{
    int l = 0,r = lx-1,mid;
    while(l<=r)
    {
        mid = (l+r)>>1;
        if(x[mid]==t) return mid;
        if(x[mid]>t) r = mid-1;
        else l = mid+1;
    }
    return -1;
}
int finy(int t)
{
    int l = 0,r = ly-1,mid;
    while(l<=r)
    {
        mid = (l+r)>>1;
        if(y[mid]==t) return mid;
        if(y[mid]>t) r = mid-1;
        else l = mid+1;
    }
    return -1;
}
int oor(int a,int b)
{
    for(int i=a;x[i]-x[a]<m;i++)
    for(int j=b;y[j]-y[b]<m;j++)
    if(v[i][j]) return 0;
    return 1;
}
int ok(int c)
{
    MM(v,0);
    REP(i,0,k)
    if(re[i].val>c)
    {
        int xl = finx(re[i].x),xr = finx(re[i].x+re[i].l);
        int yl = finy(re[i].y),yr = finy(re[i].y+re[i].l);
        REP(_i,xl,xr)
        REP(_j,yl,yr)
        v[_i][_j] = 1;
    }
    REP(i,0,lx)
    {
        if(x[i]+m>n+1) break;
        REP(j,0,ly)
        {
            if(y[j]+m>n+1) break;
            if(oor(i,j)) return 1;
        }
    }
    return 0;
}
void solve()
{
    x[lx++] = 1;y[ly++] = 1;
    x[lx++] = n+1;y[ly++] = n+1;
    REP(i,0,k)
    {
        x[lx++] = re[i].x;
        x[lx++] = re[i].x+re[i].l;
        y[ly++] = re[i].y;
        y[ly++] = re[i].y+re[i].l;
    }
    sort(x,x+lx);sort(y,y+ly);
    lx = unique(x,x+lx)-x;
    ly = unique(y,y+ly)-y;
    x[lx] = y[ly] = INF;
    int ans=-1,l=1,r=100;
    while(l<=r)
    {
        int mid = (l+r)>>1;
        if(ok(mid))
        {
            ans = mid;
            r = mid-1;
        }else
        {
            l = mid+1;
        }
    }
    if(ans==-1) puts("IMPOSSIBLE");
    else PF("%d\n",ans);
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    SI(n);SI(m);
    SI(k);
    REP(i,0,k)
    {
        re[i].get();
    }
    solve();
    return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: 
相关文章推荐