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hdu1505--City Game(扫描线+坑啊 )
City Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4572 Accepted Submission(s): 1938
Problem Description
Bob is a strategy game programming specialist. In his new city building game the gaming environment is as follows: a city is built up by areas, in which there are streets, trees,factories and buildings. There is still some space in the area that is unoccupied. The strategic task of his game is to win as much rent money from these free spaces. To win rent money you must erect buildings, that can only be rectangular, as long and wide as you can. Bob is trying to find a way to build the biggest possible building in each area. But he comes across some problems – he is not allowed to destroy already existing buildings, trees, factories and streets in the area he is building in.
Each area has its width and length. The area is divided into a grid of equal square units.The rent paid for each unit on which you‘re building stands is 3$.
Your task is to help Bob solve this problem. The whole city is divided into K areas. Each one of the areas is rectangular and has a different grid size with its own length M and width N.The existing occupied units are marked with the symbol R. The unoccupied units are marked with the symbol F.
Each area has its width and length. The area is divided into a grid of equal square units.The rent paid for each unit on which you‘re building stands is 3$.
Your task is to help Bob solve this problem. The whole city is divided into K areas. Each one of the areas is rectangular and has a different grid size with its own length M and width N.The existing occupied units are marked with the symbol R. The unoccupied units are marked with the symbol F.
Input
The first line of the input contains an integer K – determining the number of datasets. Next lines contain the area descriptions. One description is defined in the following way: The first line contains two integers-area length M<=1000 and width N<=1000, separated by a blank space. The next M lines contain N symbols that mark the reserved or free grid units,separated by a blank space. The symbols used are:
R – reserved unit
F – free unit
In the end of each area description there is a separating line.
R – reserved unit
F – free unit
In the end of each area description there is a separating line.
Output
For each data set in the input print on a separate line, on the standard output, the integer that represents the profit obtained by erecting the largest building in the area encoded by the data set.
Sample Input
25 6R F F F F FF F F F F FR R R F F FF F F F F FF F F F F F5 5R R R R RR R R R RR R R R RR R R R RR R R R R
Sample Output
450
扫描线的第一个题，题目大意：给出m*n的矩阵，R是不能占用的，F是可占用的，每占用一个F要花费3，问，占用的F组成最大的矩形时的花费
扫描线，从上向下扫描，对于每个数存下，到该点时的矩形的最大高度，矩形的左边，右边，可以通过这个计算矩形的大小，如果该点是R，那么高度是0，否则，高度+1，由该点可到的左边和上一层可到的左边共同决定该点的左边。也就是 p[i][j].l = max( p[i][j].l,p[i-1][j].l )，要求最接近该点的。
这样操作后，每个点储存的都是到给点结束时的最大矩形，找出所有点的最大值就可以了。
在输入是既要判断F 也要判断R,不然一直WA
在杭电这份代码是A的，但是其他的oj就不行了，数据水啊
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node{ int l , r , k , h ;} p[1100][1100] ;void init(int n,int m){ int i , j , l , r ; char ch[10] ; for(i = 1 ; i <= n ; i++) { for(j = 1 ; j <= m ; j++) { scanf("%s", ch); if( ch[0] == 'R' ) p[i][j].k = 0 ; else if( ch[0] == 'F' ) p[i][j].k = 1 ; } for(j = 1 , l = 1 ; j <= m ; j++) { if( p[i][j].k == 0 ) { p[i][j].l = -1 ; l = j+1 ; } else p[i][j].l = l ; } for(j = m , r = m ; j >= 1 ; j--) { if( p[i][j].k == 0 ) { p[i][j].r = 999999 ; r = j - 1 ; } else p[i][j].r = r ; } }}int main(){ int t , i , j , n , m , l , r , temp , ans ; char ch ; scanf("%d", &t); while(t--) { ans = -1 ; scanf("%d %d", &n, &m); init(n,m); for(i = 1 ; i <= m ; i++) { if( p[1][i].k == 1 ) p[1][i].h = 1 ; else p[1][i].h = 0 ; temp = p[1][i].h * ( p[1][i].r - p[1][i].l + 1); if( temp > ans ) ans = temp ; } for(i = 2 ; i <= n ; i++) { for(j = 1 ; j <= n ; j++) { if( p[i][j].k == 0 ) { p[i][j].h = 0 ; } else { p[i][j].h = p[i-1][j].h + 1 ; p[i][j].l = max( p[i][j].l,p[i-1][j].l ); p[i][j].r = min( p[i][j].r,p[i-1][j].r ); } temp = p[i][j].h * ( p[i][j].r - p[i][j].l + 1 ); if(temp > ans) ans = temp ; } } printf("%d\n", ans*3); } return 0;}
多加了一个判断是不是m行，每行是不是n个。。。。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node{ int l , r , k , h ;} p[1100][1100] ;void init(int n,int m){ int i , j , l , r , ll , top ; char str[10000] ; memset(p,0,sizeof(p)); for(i = 1 ; i <= n ; i++) { gets(str); if( str[0] == '\n' ) break; ll = strlen(str); top = 1 ; for(j = 0 ; j < ll ; j++) { if( str[j] == 'F' ) p[i][top++].k = 1 ; else if( str[j] == 'R' ) p[i][top++].k = 0 ; } } for(i = 1 ; i <= n ; i++) { for(j = 1 , l = 1 ; j <= m ; j++) { if( p[i][j].k == 0 ) { p[i][j].l = -1 ; l = j+1 ; } else p[i][j].l = l ; } for(j = m , r = m ; j >= 1 ; j--) { if( p[i][j].k == 0 ) { p[i][j].r = 999999 ; r = j - 1 ; } else p[i][j].r = r ; } }}int main(){ int t , i , j , n , m , l , r , temp , ans ; char ch ; scanf("%d", &t); while(t--) { ans = -1 ; scanf("%d %d ", &n, &m); init(n,m); for(i = 1 ; i <= m ; i++) { if( p[1][i].k == 1 ) p[1][i].h = 1 ; else p[1][i].h = 0 ; temp = p[1][i].h * ( p[1][i].r - p[1][i].l + 1); if( temp > ans ) ans = temp ; } for(i = 2 ; i <= n ; i++) { for(j = 1 ; j <= n ; j++) { if( p[i][j].k == 0 ) { p[i][j].h = 0 ; } else { p[i][j].h = p[i-1][j].h + 1 ; p[i][j].l = max( p[i][j].l,p[i-1][j].l ); p[i][j].r = min( p[i][j].r,p[i-1][j].r ); } temp = p[i][j].h * ( p[i][j].r - p[i][j].l + 1 ); if(temp > ans) ans = temp ; } } printf("%d\n", ans*3); } return 0;}
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