Minimum Sum
题目描述
One day, Snuke was given a permutation of length N, a1,a2,…,aN, from his friend.
Find the following:
Constraints
1≤N≤200,000
(a1,a2,…,aN) is a permutation of (1,2,…,N).
Find the following:
Constraints
1≤N≤200,000
(a1,a2,…,aN) is a permutation of (1,2,…,N).
输入
The input is given from Standard Input in the following format:
N
a1 a2 … aN
N
a1 a2 … aN
输出
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Note that the answer may not fit into a 32-bit integer.
样例输入
3
2 1 3
样例输出
9
题意:给出n个数,找出所有L(1~n)----R(L~n)的最小值的和。
按一般的想法来做,要用3层for循环这样肯定超时。
先找出从1-n之间的最小值,记住下标i,那么这一个最小值就会被用到(i-L+1)*(R-i+1)次,之后把这个数组从最小值哪里分成两个新的数组,之后再重复之前的过程。
这个过程和快速排序的过程类似。
然后每一次找最小值下标时如果直接找也会超时,这时候就要用到RMQ算法了。
#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll a[200105];
ll sum=0;
int d[200105][35],n;
void init()
{
int i,j;
for(i=0;i<n;i++)
d[i][0]=i;
for(j=1;(1<<j)<=n;j++)
for(i=0;i+(1<<j)-1<n;i++)
d[i][j]=a[d[i][j-1]] < a[d[i+(1<<(j-1))][j-1]]? d[i][j-1]:d[i+(1<<(j-1))][j-1];
}
int RMQ_indx_min(int s,int v)
{
int k=(int)(log((v-s+1)*1.0)/log(2.0));
return a[d[s][k]]<a[d[v-(1<<k)+1][k]]? d[s][k]:d[v-(1<<k)+1][k];
}
void dfs(int l,int r)
{
if(l==r){
sum+=(ll)a[l];
return ;
}
int minn=RMQ_indx_min(l,r);
sum=sum+(ll)(minn-l+1)*(r-minn+1)*a[minn];
if(l<=minn-1)
dfs(l,minn-1);
if(r>=minn+1)
dfs(minn+1,r);
}
int main (){
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%lld",&a[i]);
init();
dfs(0,n-1);
printf("%lld\n",sum);
return 0;
}
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