POJ 2452 (RMQ + 二分)

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#define PI acos(-1.0)
#define eps 1e-8
const int  inf =  (1<<30) - 10;
using namespace std;

const int maxx = 50000 + 10;
int n;
int a[maxx];

int dpmax[maxx][20];
int dpmin[maxx][20];
int mn[maxx];
inline void init_rmq(int a[],int n){ ///构建rmq
    mn[0]=-1;
    for(int i = 1;i <= n; ++i){
        if((i&(i-1))==0){
            mn[i] = mn[i-1]+1;
        }else{
            mn[i] = mn[i-1];
        }
        dpmax[i][0] = a[i];
        dpmin[i][0] = a[i];
    }
    for(int j = 1;j <= mn[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]);
        }
    }
}
inline int max_rmq(int l,int r){ //查询区间最大值
    int k = mn[r-l+1];
    return max(dpmax[l][k],dpmax[r-(1<<k)+1][k]);
}
inline int min_rmq(int l,int r){ //查询区间最小值
    int k = mn[r-l+1];
    return min(dpmin[l][k],dpmin[r-(1<<k)+1][k]);
}

int main(){
    while(scanf("%d",&n)==1){
        for(int i = 1;i <= n; ++i){
            scanf("%d",&a[i]);
        }
        init_rmq(a,n);
        int ans = -1;
        for(int i = 1;i <= n; ++i){
            int l = i;
            int r = n+1;
            while(l+1 != r){
                int mid = (l+r)>>1;
                if(min_rmq(i,mid) == a[i]){
                    l = mid;
                }else{
                    r = mid;
                }
            }
            int ll = i;
            int rr = l;
            if(l != i){
                int temp = max_rmq(ll,rr);
                while(ll+1 != rr){
                    int mid = (ll+rr)>>1;
                    if(max_rmq(i,mid) < temp){
                        ll = mid;
                    }else{
                        rr = mid;
                    }
                }
            }
            if(rr-i>ans) ans = rr-i;
        }
        if(ans) printf("%d\n",ans);
        else printf("-1\n");
    }
    return 0;
}

/**

7
1 2 3 7 4 5 6

*/

版权声明：本文为博主原创文章，未经博主允许不得转载。