洛谷P1966 火柴排队[NOIP提高组2013]

我确信我应该是做过这道题……就当再写一遍好了。

贪心思想，一番证明得出a和b数组中最小对最小，次小对次小……时解最优。那么先处理出a，b之间的对应关系，然后按照该关系求a或者b的逆序对数量就是答案

 /*by SilverN*/
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 using namespace std;
 const int mxn=;
 const int p=;
 struct num{
     int w;
     int num;
 }a[mxn],b[mxn];
 int cmp(num x,num y){return x.w<y.w;}
 int cmp2(num x,num y){return x.num<y.num;}
 int link[mxn];
 int t[mxn];
 int n,ans;
 int lowbit(int x){
     return x&-x;
 }
 void add(int x,int v){
     while(x<=n){
         t[x]+=v;
         x+=lowbit(x);
     }
 }
 int sum(int x){
     int res=;
     while(x){
         res+=t[x];
         x-=lowbit(x);
     }
     return res;
 }
 int main(){
     scanf("%d",&n);
     int i,j;
     for(i=;i<=n;i++)scanf("%d",&a[i].w),a[i].num=i;
     for(i=;i<=n;i++)scanf("%d",&b[i].w),b[i].num=i;
     sort(a+,a+n+,cmp);
     sort(b+,b+n+,cmp);
     for(i=;i<=n;i++) link[a[i].num]=b[i].num;
     sort(a+,a+n+,cmp2);
     for(i=;i<=n;i++){
         add(link[i],);
         ans=(ans+i-sum(link[i]))%p;
     }
     printf("%d\n",ans);
     return ;
 }