HDU2888二维RMQ
2016-06-13 00:00
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HDU2888二维RMQ
dp[row][col][i][j] 表示[row,row+2^i-1]x[col,col+2^j-1] 二维区间内的最小值
这是RMQ-ST算法的核心: 倍增思想
== min( [row,row+ 2^(i-1)-1]x[col,col+2^j-1], [row+2^(i-1),row+2^i-1]x[col,col+2^j-1] )
= min(dp[row][col][i-1][j], dp[row+(1<<(i-1))][col][i-1][j] )
//y轴不变,x轴二分 (i!=0)
或
== min( [row,row+2^i-1]x[col,col+2^(j-1)-1], [row,row+2^i-1]x[col+2^(i-1),col+2^j-1] )
= min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] )
//x轴不变,y轴二分 (j!=0)
即:
dp[row][col][i][j] = min(dp[row][col][i-1][j], dp[row + (1<<(i-1))][col][i-1][j] )
或 = min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] )
查询[x1,x2]x[y1,y2]
令 kx = (int)log2(x2-x1+1);
ky = (int)log2(y2-y1+1);
查询结果为
m1 = dp[x1][y1][kx][ky] = dp[x1][y1][kx][ky];
m2 = dp[x2-2^kx+1][y1][kx]ky] = dp[x2-(1<<kx)+1][y1][kx][ky];
m3 = dp[x1][y2-2^ky+1][kx][ky] = dp[x1][y2-(1<<ky)+1][kx][ky];
m4 = dp[x2-2^kx+1][y2-2^ky+1][kx][ky] = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];
结果 = min(m1,m2,m3,m4)
dp[row][col][i][j] 表示[row,row+2^i-1]x[col,col+2^j-1] 二维区间内的最小值
这是RMQ-ST算法的核心: 倍增思想
== min( [row,row+ 2^(i-1)-1]x[col,col+2^j-1], [row+2^(i-1),row+2^i-1]x[col,col+2^j-1] )
= min(dp[row][col][i-1][j], dp[row+(1<<(i-1))][col][i-1][j] )
//y轴不变,x轴二分 (i!=0)
或
== min( [row,row+2^i-1]x[col,col+2^(j-1)-1], [row,row+2^i-1]x[col+2^(i-1),col+2^j-1] )
= min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] )
//x轴不变,y轴二分 (j!=0)
即:
dp[row][col][i][j] = min(dp[row][col][i-1][j], dp[row + (1<<(i-1))][col][i-1][j] )
或 = min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] )
查询[x1,x2]x[y1,y2]
令 kx = (int)log2(x2-x1+1);
ky = (int)log2(y2-y1+1);
查询结果为
m1 = dp[x1][y1][kx][ky] = dp[x1][y1][kx][ky];
m2 = dp[x2-2^kx+1][y1][kx]ky] = dp[x2-(1<<kx)+1][y1][kx][ky];
m3 = dp[x1][y2-2^ky+1][kx][ky] = dp[x1][y2-(1<<ky)+1][kx][ky];
m4 = dp[x2-2^kx+1][y2-2^ky+1][kx][ky] = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];
结果 = min(m1,m2,m3,m4)
#include<iostream> #include<cmath> using namespace std; const int maxn = 301; int val[maxn][maxn]; int M,N; //RMQ 2D int dp[maxn][maxn][9][9]; void RMQ_2D_PRE() { for(int row = 1; row <= N; row++) for(int col = 1; col <=M; col++) dp[row][col][0][0] = val[row][col]; int mx = log(double(N)) / log(2.0); int my = log(double(M)) / log(2.0); for(int i=0; i<= mx; i++) { for(int j = 0; j<=my; 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 <= M; col++) { if(i == 0)//y轴二分 dp[row][col][i][j]=max(dp[row][col][i][j-1],dp[row][col+(1<<(j-1))][i][j-1]); else//x轴二分 dp[row][col][i][j]=max(dp[row][col][i-1][j],dp[row+(1<<(i-1))][col][i-1][j]); } } } } } 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 max( max(m1,m2) , max(m3,m4)); } //END int main() { int Q,x1,y1,x2,y2,i,j; while(scanf("%d%d",&N,&M)!=EOF) { for(i = 1; i <= N; i++) for(j = 1; j <= M; j++) scanf("%d",&val[i][j]); RMQ_2D_PRE(); scanf("%d",&Q); while(Q--) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); int ans = RMQ_2D(x1,x2,y1,y2); printf("%d",ans); if(ans == val[x1][y1] || ans == val[x2][y1] || ans == val[x1][y2] || ans == val[x2][y2]) printf(" yes\n"); else printf(" no\n"); } } return 0; }
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