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 = ;
 const int INF = ;

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

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

 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\n", stx, sty, edx, edy );
     //printf("%d %d %d %d\n---\n", 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 = ; i < cntX; ++i )
         for( int j = ; j < cntY; ++j )
         {
             Land park;
             park.x = X[i], park.y = Y[j];
             park.len = K;

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

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

     return;
 }

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

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

         solved();
     }
     return ;
 }