URAL 1097 Square Country 2 离散化

一共才100个正方形，将所有正方形左下角和右上角的X坐标和Y坐标离散化，直接枚举新建公园的点的坐标即可。

O(n^3)的时间复杂度。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int MAXN = 110;
const int INF = 255;

struct Land
{
    int x, y;
    int len;
    int import;
};

int N, K, M;
int cnt;
Land land[MAXN];
int cntX, cntY;
int X[ MAXN << 1 ];
int Y[ MAXN << 1 ];

bool InPark( Land park, Land d )
{
    int stx = park.x, sty = park.y;
    int edx = park.x + park.len, edy = park.y + park.len;

    //printf("%d %d %d %d
", stx, sty, edx, edy );
    //printf("%d %d %d %d
---
", d.x, d.y, d.x + d.len, d.y + d.len );

    if ( d.x >= stx && d.x < edx )
    {
        if ( d.y >= sty && d.y < edy ) return true;
        if ( d.y + d.len > sty && d.y + d.len <= edy ) return true;
    }

    if ( d.x + d.len > stx && d.x + d.len <= edx )
    {
        if ( d.y >= sty && d.y < edy ) return true;
        if ( d.y + d.len > sty && d.y + d.len <= edy ) return true;
    }
    return false;
}

void solved()
{
    int ans = INF;
    for( int i = 0; i < cntX; ++i )
        for( int j = 0; j < cntY; ++j )
        {
            Land park;
            park.x = X[i], park.y = Y[j];
            park.len = K;

            if ( park.x + park.len <= 1 + N && park.y + park.len <= 1 + N )
            {
                int influence = 1;
                for ( int k = 0; k < M; ++k )
                {
                    if ( InPark( park, land[k] ) )
                    {
                        //puts("***");
                        influence = max( influence, land[k].import );
                    }
                }
                ans = min( ans, influence );
            }
        }

    if ( ans <= 100 ) printf( "%d
", ans );
    else puts("IMPOSSIBLE");

    return;
}

int main()
{
    //freopen( "s.out", "w", stdout );
    while ( ~scanf( "%d%d", &N, &K ) )
    {
        scanf( "%d", &M );
        cntX = cntY = 0;
        for ( int i = 0; i < M; ++i )
        {
            scanf("%d%d%d%d", &land[i].import, &land[i].len, &land[i].x, &land[i].y );
            X[cntX++] = land[i].x;
            X[cntX++] = land[i].x + land[i].len;
            Y[cntY++] = land[i].y;
            Y[cntY++] = land[i].y + land[i].len;
        }

        X[cntX++] = 1;
        X[cntX++] = 1 + N;
        Y[cntY++] = 1;
        Y[cntY++] = 1 + N;

        sort( X, X + cntX );
        sort( Y, Y + cntY );
        cntX = unique( X, X + cntX ) - X;
        cntY = unique( Y, Y + cntY ) - Y;

        solved();
    }
    return 0;
}