HDU-2888 Check Corners 二维RMQ
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2888
模板题。解题思路如下（转载别人写的)：
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)
1 //STATUS:C++_AC_4109MS_30160KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef long long LL; 34 typedef unsigned long long ULL; 35 //const 36 const int N=310; 37 const int INF=0x3f3f3f3f; 38 const int MOD=1e+7,STA=8000010; 39 const LL LNF=1LL<<60; 40 const double EPS=1e-8; 41 const double OO=1e15; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 int val[N][N],dp[N][N][9][9]; 58 int n,m,Q; 59 void init() 60 { 61 for(int row = 1; row <= n; row++) 62 for(int col = 1; col <=m; col++) 63 dp[row][col][0][0] = val[row][col]; 64 int mx = log(double(n)) / log(2.0); 65 int my = log(double(m)) / log(2.0); 66 for(int i=0; i<= mx; i++) 67 { 68 for(int j = 0; j<=my; j++) 69 { 70 if(i == 0 && j ==0) continue; 71 for(int row = 1; row+(1<<i)-1 <= n; row++) 72 { 73 for(int col = 1; col+(1<<j)-1 <= m; col++) 74 { 75 if(i == 0)//y轴二分 76 dp[row][col][i][j]=max(dp[row][col][i][j-1],dp[row][col+(1<<(j-1))][i][j-1]); 77 else//x轴二分 78 dp[row][col][i][j]=max(dp[row][col][i-1][j],dp[row+(1<<(i-1))][col][i-1][j]); 79 } 80 } 81 } 82 } 83 } 84 int RMQ2D(int x1,int y1,int x2,int y2) 85 { 86 int kx = log(double(x2-x1+1)) / log(2.0); 87 int ky = log(double(y2-y1+1)) / log(2.0); 88 int m1 = dp[x1][y1][kx][ky]; 89 int m2 = dp[x2-(1<<kx)+1][y1][kx][ky]; 90 int m3 = dp[x1][y2-(1<<ky)+1][kx][ky]; 91 int m4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky]; 92 return max( max(m1,m2) , max(m3,m4)); 93 } 94 //END 95 int main() 96 { 97 // freopen("in.txt","r",stdin); 98 int i,j,ans; 99 int x1,y1,x2,y2; 100 while(~scanf("%d%d",&n,&m)) 101 { 102 for(i=1;i<=n;i++) 103 for(j=1;j<=m;j++) 104 scanf("%d",&val[i][j]); 105 init(); 106 scanf("%d",&Q); 107 while(Q--){ 108 scanf("%d%d%d%d",&x1,&y1,&x2,&y2); 109 ans=RMQ2D(x1,y1,x2,y2); 110 if(ans==val[x1][y1] || ans==val[x2][y2] 111 || ans==val[x1][y2] || ans==val[x2][y1]){ 112 printf("%d yes\n",ans); 113 } 114 else printf("%d no\n",ans); 115 } 116 } 117 return 0; 118 }
