1348：数组中的逆序对（树状数组 or 归并）

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OJ题目：click here~~

题目分析：经典题目，求逆序对。两种方法：

一、离散化+树状数组

二、归并排序

利用归并排序，已A：

AC_CODE

#include<stdio.h>#include<string.h>using namespace std ;typedef long long LL ;const int maxn = 100008 ;int n ;LL ans ;int x[maxn] ;void Merge(int low , int high){    int mid = (low + high)/2 ;    int i = low, j = mid + 1, k = 0 ;    int xx[maxn] ;    while(i <= mid && j <= high){        if(x[i] > x[j]){            xx[k++] = x[j++] ;            ans += (mid - i + 1) ;        }        else{            xx[k++] = x[i++] ;        }    }    if(j <= high) while(j <= high) xx[k++] = x[j++] ;     if(i <= mid) while(i <= mid) xx[k++] = x[i++] ;    k = 0 ;    i = low ;    while(i <= high) x[i++] = xx[k++] ;} void MergSort(int low , int high){    if(high <= low) return ;    int mid = (low + high)/2 ;    MergSort(low , mid) ;    MergSort(mid + 1 , high) ;    Merge(low , high) ;} int main(){    int i , j , k ;    while(scanf("%d",&n) != EOF){        ans = 0 ;        for(i = 0;i < n;i++)            scanf("%d",&x[i]) ;        MergSort(0 , n - 1) ;        printf("%lld\n",ans) ;    }    return 0 ;}

树状数组，不能A，但是在POJ2299能A。。。。。

#include <stdio.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <math.h>#include <string.h>using namespace std;typedef long long LL ;const int maxn = 100005 ; int n ;int x[maxn] ;LL c[maxn] ; struct Node{    int x,order ;}node[maxn]; bool cmp(const Node &A ,const Node &B){    return A.x < B.x ;} int lowbit(int x){    return x&(-x) ;} void update(int t , int x){    int i ;    for(i = t;i <= n;i += lowbit(i)){        c[i] += x ;    }} int getsum(int x){    int i ,sum = 0 ;    for(i = x;i > 0;i -= lowbit(i)){        sum += c[i] ;    }    return sum ;} int main(){    int i ;    while(scanf("%d",&n) != EOF){        memset(c , 0 , sizeof(c)) ;        for (i = 1;i <= n;i++){            scanf("%d",&node[i].x) ;            node[i].order = i ;        }        sort(node + 1 , node + 1 + n , cmp) ;        for(i = 1;i <= n;i++)            x[node[i].order] = i ;        LL sum = 0 ;        for(i = 1;i <= n;i++){            update(x[i] , 1) ;            sum += (i - getsum(x[i])) ;        }        cout << sum << endl ;    }    return 0;} /**************************************************************    Problem: 1348    Language: C++    Result: Wrong Answer****************************************************************/

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