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题意是 求最小的矩形覆盖面积内包含 k 个 空位置
枚举上下边界然后 双端队列 求 最小面积
#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <map> #include <queue> #include <stack> #include <set> #include <string> using namespace std; typedef long long ll; const double ESP = 10e-8; const int MOD = 1000000000+7; const int MAXN = 300+10; char graph[MAXN][MAXN]; int sum[MAXN][MAXN]; int main(){ // freopen("input.txt","r",stdin); int R,C,K; while(scanf("%d%d%d",&R,&C,&K)){ if(!R && !C && !K){ break; } for(int i = 0;i < R;i++){ scanf("%s",graph[i]); } memset(sum,0,sizeof(sum)); for(int i = 1;i <= R;i++){ for(int j = 1;j <= C;j++){ sum[i][j] = sum[i][j-1]; sum[i][j] += sum[i-1][j] - sum[i-1][j-1]; if(graph[i-1][j-1] == ‘.‘){ sum[i][j]++; } } } int ans = R * C; for(int x2 = R;x2 > 0;x2--){ if(sum[x2][C] < K){ break; } for(int x1 = 1;x1 <= R;x1++){ if(sum[x2][C] - sum[x1-1][C] < K){ break; } int y1 = 1; int y2 = 1; while(y1 <= C && y2 <= C){ int cnt = sum[x2][y2]-sum[x1-1][y2]- (sum[x2][y1-1] - sum[x1-1][y1-1]); if(cnt < K){ y2++; }else{ ans = min(ans,(x2-x1+1)*(y2-y1+1)); y1++; if(y1 > y2){ break; } } } } } printf("%d\n",ans); } return 0; }
跟上面是类似的题,今年省赛的题 T_T
求最小矩形面积覆盖的 星星数 至少 为 k 个
就跟上面一样的题型了23333
#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <map> #include <queue> #include <stack> #include <set> #include <string> using namespace std; typedef long long ll; const double ESP = 10e-8; const int MOD = 1000000000+7; const int MAXN = 400+10; int graph[MAXN][MAXN]; int sum[MAXN][MAXN]; int main(){ //freopen("input.txt","r",stdin); int t; scanf("%d",&t); while(t--){ int n,k; scanf("%d%d",&n,&k); memset(graph,0,sizeof(graph)); memset(sum,0,sizeof(sum)); int stX = 400,edX = 1,stY = 400,edY = 1; while(n--){ int a,b; scanf("%d%d",&a,&b); graph[a][b]++; stX = min(a,stX); stY = min(b,stY); edX = max(a,edX); edY = max(b,edY); } for(int i = stX;i <= edX;i++){ for(int j = stY;j <= edY;j++){ sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1]; sum[i][j] += graph[i][j]; } } int ans = (edX-stX+1)*(edY-stY+1); for(int x2 = edX;x2 >= stX;x2--){ if(sum[x2][edY] < k){ break; } for(int x1 = stX;x1 <= edX;x1++){ if(sum[x2][edY] - sum[x1-1][edY] < k){ break; } int y1 = stY; int y2 = stY; while(y1 <= edY && y2 <= edY){ int cnt = sum[x2][y2] - sum[x2][y1-1]-(sum[x1-1][y2]-sum[x1-1][y1-1]); if(cnt < k){ y2++; }else{ ans = min(ans,(x2-x1+1)*(y2-y1+1)); y1++; if(y1 > y2){ break; } } } } } printf("%d\n",ans); } return 0; }
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原文地址:http://www.cnblogs.com/hanbinggan/p/4678832.html