标签：style class blog code color get
题目描述
Given a non-negative integer sequence A with length N, you can exchange two adjacent numbers each time. After K exchanging operations, what’s the minimum reverse number the sequence can achieve? The reverse number of a sequence is the number of pairs (i, j) such that i < j and Ai > Aj
输
入There are multiple cases. For each case, first line contains two numbers: N, K 2<=N<=100000, 0 <= K < 2^60 Second line contains N non-negative numbers, each of which not greater than 2^30
输出
Minimum reverse number you can get after K exchanging operations.
样例输入
3 1 3 2 1 5 2 5 1 4 3 2
样例输出
Case #1: 2 Case #2: 5
本题当然是先求逆序对，归并排序，树状数组，线段树，什么都可以，主要是先求出来，做k次变化，我们的目标不就是找刚好是逆序的变换，这样再看看k是否大于逆序对数的情况，不就完了
#include<map> #include<set> #include<queue> #include<cmath> #include<vector> #include<cstdio> #include<string> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define inf 0x0f0f0f0f using namespace std; const int maxn=100000+10; int a[maxn],b[maxn],ans; void merge_sort(int* A,int x,int y,int*T) { if (y-x>1) { int m=x+(y-x)/2; int p=x, q=m, i=x; merge_sort(A,x,m,T); merge_sort(A,m,y,T); while (p<m || q<y) { if (q>=y || (p<m && A[p]<=A[q])) T[i++]=A[p++]; else {T[i++]=A[q++];ans+=m-p;} } for (int i=x;i<y;i++) A[i]=T[i]; } } int main() { int n,k,t=0; while(scanf("%d%d",&n,&k)!=EOF) { t++; for (int i=0;i<n;i++) scanf("%d",&a[i]); ans=0; merge_sort(a,0,n,b); ans-=k; bool cut=false; for (int i=0;i<n-1;i++) if (a[i]==a[i+1]) {cut=true;break;} if (ans<0) { if ((-ans)%2==0) ans=0; else { if(cut) ans=0; else ans=1; } } printf("Case #%d: %d\n",t,ans); } return 0; }
作者 chensunrise
hust 1347 - Reverse Number,布布扣,bubuko.com
标签：style class blog code color get
原文地址：http://www.cnblogs.com/chensunrise/p/3800102.html
