P1966 火柴排队

Posted 2021-05-07 hahaha2124652975

emmmmm不难吧。。

主要就是在最开始要把题目抽象化，转为求逆序对个数，那之后就很简单了，离散化+归并排序求逆序对，取模输出就结束了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<queue>
 6 #include<stack>
 7 #include<deque>
 8 #include<algorithm>
 9 #define ll long long
10 using namespace std;
11 const int oo=0x3f3f3f3f;
12 const int N=1000005;
13 const int mod=99999997;
14 
15 int n,ans;
16 int f[N],f1[N];
17 struct node{
18     int x,y;
19 }a[N];
20 struct node1{
21     int x,y;
22 }b[N];
23 
24 bool cmp1(node a,node b){return a.x<b.x;}
25 bool cmp2(node1 a,node1 b){return a.x<b.x;}
26 
27 ll get(){
28     char zy=getchar();
29     ll z=1,y=0;
30     while(zy>‘9‘||zy<‘0‘){
31         if(zy==‘-‘) z=-1;
32         zy=getchar();
33     }
34     while(zy>=‘0‘&&zy<=‘9‘){
35         y=(y<<1)+(y<<3)+zy-‘0‘;
36         zy=getchar();
37     }
38     return z*y;
39 }
40 
41 void merge_sort(int l,int r){
42     if(l>=r) return;
43     int mid=(l+r)>>1;
44     merge_sort(l,mid);
45     merge_sort(mid+1,r);
46     int p=l,q=mid+1,g=l;
47     while(p<=mid||q<=r){
48         if(q>r||(f[p]<=f[q]&&p<=mid)) f1[g++]=f[p++];
49         else{
50             ans+=mid-p+1;
51             ans%=mod;
52             f1[g++]=f[q++];
53         }
54     }
55     for(int i=l;i<=r;i++){
56         f[i]=f1[i];
57     }
58 }
59 
60 int main(){
61     //freopen("P1966火柴排队.in","r",stdin);
62     //freopen("P1966火柴排队.out","w",stdout);
63     n=get();
64     for(int i=1;i<=n;i++){
65         a[i].x=get();a[i].y=i;
66     }
67     sort(a+1,a+1+n,cmp1);
68     for(int i=1;i<=n;i++){
69         b[i].x=get();b[i].y=i;
70     }
71     sort(b+1,b+1+n,cmp2);
72     for(int i=1;i<=n;i++){
73         f[a[i].y]=b[i].y;
74     }
75     for(int i=1;i<=n;i++){
76     //    printf("%d ",f[i]);
77     }
78     merge_sort(1,n);
79     printf("%d
",ans%mod);
80     return 0;
81 }