poj 2299 逆序数树状数组

（1)用结构体存每个数的数置和它的值，struct cnode{int value,int pos}; 

(2)pos位置取负再加上ＭＡＸＮ＋１，ＭＡＸＮ（是元素位置的最大值），

（３）排序：先按value从小到大排，再按pos从小到大排，剩下的就是求每个元素左下角的元素个数，交给树状数组来处理是再好不过的事了。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int MAXN=500000;struct cnode{ int pos; int value;}node[MAXN+10];int tree[MAXN+10];bool cmp(const cnode &e1,const cnode &e2){ if(e1.value!=e2.value) { return e1.value<e2.value? true:false; } else { return e1.pos<e2.pos? true:false; }}int n;int sum(int x){ int value=0; while(x>0) { value+=tree[x]; x-=(x&(-x)); } return value;}void insert(int x){ while(x<=MAXN) { tree[x]++; x+=(x&(-x)); }}int main(){ //int t; //freopen("tree.out","w",stdout); // freopen("1804.in","r",stdin); // scanf("%d",&t); int num[MAXN+10]; int sub; int i; int Case=0; // while(t--) // { while(scanf("%d",&n)!=EOF) { if(n==0){ break;} memset(tree,0,sizeof(tree)); Case++; // scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%d",&num[i]); } sub=1; for(i=n;i>0;i--){ node[sub].pos=-i+MAXN+1; node[sub].value=num[i]; sub++; } sort(&node[1],&node[n+1],cmp); long long ans=0; int samenum=0; node[0].value=-1000000-1; for(i=1;i<=n;i++){ if(node[i].value!=node[i-1].value) { samenum=0; ans+=sum(node[i].pos-1); //insert(node[i].pos); //ans+=beforevalue; } else { samenum++; ans+=sum(node[i].pos-1)-samenum; // insert(node[i].pos); } insert(node[i].pos); } // printf("Scenario #%d:/n%d/n/n",Case,ans); printf("%lld/n",ans); } return 0;}

题目：

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 20199 Accepted: 7146

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

59105431230Sample Output60Source

Waterloo local 2005.02.05