dp+离散（RMQ）

一、RMQ问题描述

RMQ (Range Minimum/Maximum Query)问题是指：对于长度为n的数列A，回答若干询问RMQ(A,i,j)(i,j<=n)，返回数列A中下标在[i,j]里的最小(大）值，也就是说，RMQ问题是指求区间最值的问题

ST算法（Sparse Table）:它是一种动态规划的方法。 
以最小值为例。a为所寻找的数组. 
用一个二维数组dp(i,j)记录区间[i,i+2^j-1](持续2^j个)区间中的最小值。其中dp[i,0] = a[i]; 所以，对于任意的一组(i,j)，dp(i,j) = min{dp(i,j-1),dp(i+2^(j-1),j-1)}来使用动态规划计算出来。 
这个算法的高明之处不是在于这个动态规划的建立，而是它的查询：它的查询效率是O(log(j-i)). 假设我们要求区间[m,n]中a的最小值，找到一个数k使得2^k<n-m+1)并且2^(k+1)>=n-m+1. 
这样，可以把这个区间分成两个部分：[m,m+2^k-1]和[n-2^k+1,n].这两个部分要一定保证完全覆盖这个区间，我们发现，这两个区间是已经初始化好的. 
前面的区间是dp(m,k)，后面的区间是dp(n-2^k+1,k). 
这样，只要看这两个区间的最小值，就可以知道整个区间的最小值！

二、RMQ的应用

1、poj1207

/*裸的RMQ*/#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define max(a,b) ((a)>(b)?(a):(b))int dat[10002];int dp[10002][18];int get_dat(int n){int cnt=1;while(n!=1){if(n&1){n=3*n+1;}else{n/=2;}++cnt;}return cnt;}void make_max_rmq(){for(int j=1;(1<<j)<=10000;++j){for(int i=1;i+(1<<j)-1<=10000;++i){dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}}}int get_max_rmq(int a,int b){int k=0;while(a+(1<<k)<b-(1<<k)+1)++k;return max(dp[a][k],dp[b-(1<<k)+1][k]);}int main(){for(int i=1;i<=10000;++i){dat[i]=get_dat(i);dp[i][0]=dat[i];}int l,h,a,b;make_max_rmq();while(~scanf("%d%d",&a,&b)){l=a<b?a:b;h=a>b?a:b;int ans=get_max_rmq(l,h);printf("%d %d %d\n",a,b,ans);}return 0;}

 

2、poj3264

/*n个数，求i，j的最大值和最小值之差 */#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))int m,n;int dat[50002];int dpmax[50002][18];int dpmin[50002][18];void make_max_rmq(){for(int i=1;i<=n;++i){dpmax[i][0]=dat[i];}for(int j=1;(1<<j)<=n;++j){for(int i=1;i+(1<<j)-1<=n;++i){dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]);}}}void make_min_rmq(){for(int i=1;i<=n;++i){dpmin[i][0]=dat[i];}for(int j=1;(1<<j)<=n;++j){for(int i=1;i+(1<<j)-1<=n;++i){dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]);}}}int get_max_rmq(int a,int b){int k=0;while(a+(1<<k)<b-(1<<k)+1)++k;return max(dpmax[a][k],dpmax[b-(1<<k)+1][k]);}int get_min_rmq(int a,int b){int k=0;while(a+(1<<k)<b-(1<<k)+1)++k;return min(dpmin[a][k],dpmin[b-(1<<k)+1][k]);}int main(){int a,b;while(~scanf("%d%d",&n,&m)){for(int i=1;i<=n;++i){scanf("%d",&dat[i]);}make_max_rmq();make_min_rmq();while(m--){scanf("%d%d",&a,&b);int ans=get_max_rmq(a,b)-get_min_rmq(a,b);printf("%d\n",ans);}}return 0;}

 

3、poj3368

/* * 有n个数升序排列，输出i,j的最大的众数 */#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;#define max(a,b) ((a)>(b)?(a):(b))#define N 100002int n,m;int rank[N];int dat[N];int dp[N][18];int rear[N];int head[N];void make_max_rmq(){for(int i=1;i<=n;++i){dp[i][0]=rank[i];}for(int j=1;j<=(int)(log((double)n)/log(2.0));++j){for(int i=1;i+(1<<j)-1<=n;++i){dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}}}int get_max_rmq(int a,int b){int k=0;while(a+(1<<k)<b-(1<<k)+1)++k;  return max(dp[a][k],dp[b-(1<<k)+1][k]);}int main(){int a,b;while(~scanf("%d",&n)){if(!n)break;scanf("%d",&m);int cnt=0;memset(head,0,sizeof(head));memset(rear,0,sizeof(rear));head[0]=1;for(int i=1;i<=n;++i){scanf("%d",&dat[i]);if(dat[i]==dat[i-1]||i==1){rank[i]=++cnt;head[i]=head[i-1];}else{head[i]=i;rear[head[i-1]]=i-1;rank[i]=1;cnt=1;}}make_max_rmq();while(m--){scanf("%d%d",&a,&b);int ans;if(dat[a]==dat[b]){ans=b-a+1;}else if(rear[head[a]]+1==head[b]){ans=max(rear[head[a]]-a+1,b-head[b]+1);}else{ans=max(rear[head[a]]-a+1,b-head[b]+1);ans=max(ans,get_max_rmq(rear[head[a]]+1,head[b]-1));}printf("%d\n",ans);}}return 0;}

 

4、poj2109 二维RMQ

/* 求子矩阵的最大值与最小值 */#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))#define inf 0x7ffffffint dpmax[252][252][8];int dpmin[252][252][8];int dat[252][252];int n,b,k;void make_rmq2(){for(int k=1;(1<<k)<=n;++k){for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){dpmax[i][j][k]=dpmax[i][j][k-1];if(i+(1<<(k-1))<=n)dpmax[i][j][k]=max(dpmax[i][j][k],dpmax[i+(1<<(k-1))][j][k-1]);if(j+(1<<(k-1))<=n)dpmax[i][j][k]=max(dpmax[i][j][k],dpmax[i][j+(1<<(k-1))][k-1]);if((i+(1<<(k-1))<=n)&&(j+(1<<(k-1))<=n))dpmax[i][j][k]=max(dpmax[i][j][k],dpmax[i+(1<<(k-1))][j+(1<<(k-1))][k-1]);dpmin[i][j][k]=dpmin[i][j][k-1];if(i+(1<<(k-1))<=n)dpmin[i][j][k]=min(dpmin[i][j][k],dpmin[i+(1<<(k-1))][j][k-1]);if(j+(1<<(k-1))<=n)dpmin[i][j][k]=min(dpmin[i][j][k],dpmin[i][j+(1<<(k-1))][k-1]);if((i+(1<<(k-1))<=n)&&(j+(1<<(k-1))<=n))dpmin[i][j][k]=min(dpmin[i][j][k],dpmin[i+(1<<(k-1))][j+(1<<(k-1))][k-1]);}}}}int get_max_rmq2(int x,int y,int k){int r=dpmax[x][y][k];if(x+b-(1<<k)<=n)r=max(r,dpmax[x+b-(1<<k)][y][k]);if(y+b-(1<<k)<=n)r=max(r,dpmax[x][y+b-(1<<k)][k]);if((x+b-(1<<k)<=n)&&(y+b-(1<<k)<=n))r=max(r,dpmax[x+b-(1<<k)][y+b-(1<<k)][k]);return r;}int get_min_rmq2(int x,int y,int k){int r=dpmin[x][y][k];if(x+b-(1<<k)<=n)r=min(r,dpmin[x+b-(1<<k)][y][k]);if(y+b-(1<<k)<=n)r=min(r,dpmin[x][y+b-(1<<k)][k]);if((x+b-(1<<k)<=n)&&(y+b-(1<<k)<=n))r=min(r,dpmin[x+b-(1<<k)][y+b-(1<<k)][k]);return r;}int main(){int x,y;while(~scanf("%d%d%d",&n,&b,&k)){for(int i=1;i<=n;++i){for(int j=1;j<=n;++j){scanf("%d",&dat[i][j]);dpmax[i][j][0]=dpmin[i][j][0]=dat[i][j];}}int p=0;while((1<<(p+1))<=b){++p;}make_rmq2();while(k--){scanf("%d%d",&x,&y);int ans=get_max_rmq2(x,y,p)-get_min_rmq2(x,y,p);printf("%d\n",ans);}}return 0;}

 

5、Poj2452 RMQ+两次二分

/*题意：有一串N个不同数字的序列，求出一个最长的连续子序列si~sj，满足si是最小数，sj是最大数。思路：首先预处理，对于每个si,找出以它为起点后面最大数的位置rear[i].然后二分找到以si为最小值的区间，然后在这个区间里面二分寻找最大值的位置j，j-i的最大值即为所求*/#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))。const int N=5e4+2;int n;int dat[N];int dpmax[N][30];int dpmin[N][30];int rear[N];void make_rmq(){for(int j=1;(1<<j)<=n;++j){for(int i=1;i+(1<<j)-1<=n;++i){dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]);dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]);}}}int get_max_rmq(int a,int b){int k=0;while(a+(1<<k)<b-(1<<k)+1)++k;return max(dpmax[a][k],dpmax[b-(1<<k)+1][k]);}int get_min_rmq(int a,int b){int k=0;while(a+(1<<k)<b-(1<<k)+1)++k;return min(dpmin[a][k],dpmin[b-(1<<k)+1][k]);}int main(){while(~scanf("%d",&n)){for(int i=1;i<=n;++i){scanf("%d",&dat[i]);dpmin[i][0]=dpmax[i][0]=dat[i];}int t=n;rear[n]=n;for(int i=n-1;i>=1;--i){if(dat[i]<dat[t]){rear[i]=t;}else{rear[i]=i;t=i;}}for(int i=1;i<=n;++i){}int ans=-1;make_rmq();for(int i=1;i<=n;++i){int low=i,high=rear[i],mid,t=-1;if(get_min_rmq(i,high)==dat[i]){t=max(t,high);}else{while(low<high){if(get_min_rmq(i,low)==dat[i]){t=max(t,low);}if(get_min_rmq(i,high)==dat[i]){t=max(t,high);}mid=(low+high)/2;if(get_min_rmq(i,mid)==dat[i]){t=max(t,mid);low=mid+1;}else{high=mid-1;}}}high=t;low=i;int w=get_max_rmq(low,high);if(get_max_rmq(i,high)==dat[high]){ans=max(ans,high-i);}else{while(low<high){if(dat[low]==w){ans=max(ans,low-i);}if(dat[high]==w){ans=max(ans,high-i);}mid=(low+high)/2;if(get_max_rmq(i,mid)<w){low=mid+1;}else{high=mid-1;if(dat[mid]==w)ans=max(ans,mid-i);}}}}if(ans==0)ans=-1;printf("%d\n",ans);}return 0;}