HDU 1754 I Hate It 线段树

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13207    Accepted Submission(s): 8069

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

Output

For each case, output the minimum inversion number on a single line.

 

Sample Input

10
1 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

Author

CHEN, Gaoli

 

个人写的第二个线段树程序

更新单点  询问区间最大值

#include <iostream>
#include <cstdio>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#define esp 1e-6
#define inf 0x0f0f0f0f
#define LL long long  
using namespace std;


#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int maxn=222222;
int sum[maxn<<2];

void PushUp(int rt){
    sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);
}

void build(int l,int r,int rt){
    if(l==r){
        scanf("%d",&sum[rt]);
        return;
    }
    int m=(r+l)>>1;
    build(lson);
    build(rson);
    PushUp(rt);
}

void update(int p,int add,int l,int r,int rt){
    if(l==r){
        sum[rt]=add;
        return;
    }
    int m=(l+r)>>1;
    if(p<=m) update(p,add,lson);
    else update(p,add,rson);
    PushUp(rt);    
}

int query(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R )
        return sum[rt];
    int m=(l+r)>>1;
    int ret=0;
    if(L<=m)ret=max(ret,query(L,R,lson));
    if(R>m) ret=max(ret,query(L,R,rson));
    return ret;
}

int main(){
    int T,n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
    	memset(sum,0,sizeof(sum));
        build(1,n,1);
        char op[10];
        for(int i=0;i<m;i++){
        	scanf("%s",op);
            int a,b;
            scanf("%d%d",&a,&b);
            if(op[0]=='Q') printf("%d\n",query(a,b,1,n,1));
            else update(a,b,1,n,1);
/*
            for(int j=0;j<20;j++){
            	printf("%d ",sum[j]);
            }*/
    	}

    }
}