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

 

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=120000;
 9 const int p=99999997;
10 struct num{
11     int w;
12     int num;
13 }a[mxn],b[mxn];
14 int cmp(num x,num y){return x.w<y.w;}
15 int cmp2(num x,num y){return x.num<y.num;}
16 int link[mxn];
17 int t[mxn];
18 int n,ans;
19 int lowbit(int x){
20     return x&-x;
21 }
22 void add(int x,int v){
23     while(x<=n){
24         t[x]+=v;
25         x+=lowbit(x);
26     }
27 }
28 int sum(int x){
29     int res=0;
30     while(x){
31         res+=t[x];
32         x-=lowbit(x);
33     }
34     return res;
35 }
36 int main(){
37     scanf("%d",&n);
38     int i,j;
39     for(i=1;i<=n;i++)scanf("%d",&a[i].w),a[i].num=i;
40     for(i=1;i<=n;i++)scanf("%d",&b[i].w),b[i].num=i;
41     sort(a+1,a+n+1,cmp);
42     sort(b+1,b+n+1,cmp);
43     for(i=1;i<=n;i++) link[a[i].num]=b[i].num;
44     sort(a+1,a+n+1,cmp2);
45     for(i=1;i<=n;i++){
46         add(link[i],1);
47         ans=(ans+i-sum(link[i]))%p;
48     }
49     printf("%d\n",ans);
50     return 0;
51 }

 

posted @ 2016-08-10 21:27  SilverNebula  阅读(160)  评论(0)  编辑  收藏  举报
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