标签:blog os io for 2014 时间 amp log
这题求范围最小值,RMQ正好是用来解决这方面的,所以再适合不过了,又是离线静态输入输出的,所以时间比二维线段树快。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<map> #include<queue> #include<set> #include<cmath> #include<bitset> #define mem(a,b) memset(a,b,sizeof(a)) #define lson i<<1,l,mid #define rson i<<1|1,mid+1,r #define INF 510010 #define maxn 310 using namespace std; typedef long long ll; typedef unsigned long long ull; int N; int val[301][301]; int dp[maxn][maxn][9][9]; void RMQ_2D() { for(int row = 1; row <= N; row++) for(int col = 1; col<=N; col++) dp[row][col][0][0] = val[row][col]; int m = log(double(N)) / log(2.0); for(int i=0; i<=m; i++) for(int j=0; j<=m; j++) { if(i == 0 && j==0) continue; for(int row = 1; row+(1<<i)-1 <= N; row++) for(int col = 1; col+(1<<j)-1 <= N; col++) { if(i == 0) dp[row][col][i][j] = min(dp[row][col][i][j-1] , dp[row][col+(1<<(j-1))][i][j-1]); //水平划分 else dp[row][col][i][j] = min(dp[row][col][i-1][j] , dp[row+(1<<(i-1))][col][i-1][j]); //竖直划分 } } } int RMQ_2D(int x1,int x2,int y1,int y2) { int kx = log(double(x2 - x1 +1)) / log(2.0); int ky = log(double(y2 - y1 +1)) / log(2.0); int m1 = dp[x1][y1][kx][ky]; int m2 = dp[x2-(1<<kx)+1][y1][kx][ky]; int m3 = dp[x1][y2-(1<<ky)+1][kx][ky]; int m4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky]; return min( min(m1,m2), min(m3,m4) ); } int main() { int T; scanf("%d",&T); int M; int x1,y1,x2,y2; while(T--) { scanf("%d",&N); for(int i=1; i<=N; i++) for(int j=1; j<=N; j++) scanf("%d",&val[i][j]); RMQ_2D(); scanf("%d",&M); while(M--) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); printf("%d\n",RMQ_2D(x1,x2,y1,y2)); } } return 0; }
ZOJ 2859 二维RMQ(模板),布布扣,bubuko.com
标签:blog os io for 2014 时间 amp log
原文地址:http://blog.csdn.net/u011466175/article/details/38456217