LA 3029(悬线法)
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City Game
64-bit integer IO format: %lld Java class name: Main
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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.
Input
R reserved unitIn the end of each area description there is a separating line.
F free unit
Output
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
Source
#include <iostream>#include <cstdio>#include <cstring>#include <stack>using namespace std;typedef long long ll;const int maxn = 1e4 + 10;int a[maxn],s[maxn][maxn];int main(){ int T;scanf("%d",&T); while(T--) { int n,m;scanf("%d%d",&n,&m); stack<int> Q; memset(a,0,sizeof a); int ans = 0; for(int i = 1; i <= n; ++i) { while(!Q.empty()) Q.pop(); Q.push(0); a[0] = -1; for(int j = 1; j <= m; ++j) { ++a[j]; char ch = getchar(); while(ch != 'R' && ch != 'F') ch = getchar(); s[i][m-j+1] = ch; if(ch == 'R') a[j] = 0; while(a[Q.top()] >= a[j]) Q.pop(); ans = max(ans,(j-Q.top())*a[j]); Q.push(j); } } memset(a,0,sizeof a); for(int i = 1; i <= n; ++i) { while(!Q.empty()) Q.pop(); Q.push(0); a[0] = -1; for(int j = 1; j <= m; ++j) { ++a[j]; char ch = s[i][j]; if(ch == 'R') a[j] = 0; while(a[Q.top()] >= a[j]) Q.pop(); ans = max(ans,(j-Q.top())*a[j]); Q.push(j); } } printf("%d\n",ans*3); } return 0;}
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