逆序对,树状数组,归并排序

求逆序对有三种方法,

线段树求逆序对,

树状数组求逆序对,

分治求逆序对,

 

先上手树状数组版的,

有没有重复的数都好使,

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <string>
 4 #include <bits/stdc++.h>
 5 
 6 using namespace std;
 7 typedef long long ll;
 8 typedef unsigned long long ull;
 9 
10 #define MAXN 500001
11 ll c[MAXN];
12 //传说中的树状数组 
13 ll n;
14 struct shu{
15     ll val;
16     ll pos;
17 }num[MAXN];
18 bool cmp(shu a,shu b) {
19     //这种是有重复数的 
20     if (a.val!=b.val) return a.val>b.val;
21     else return a.pos>b.pos; 
22     /* retuen a.val>b.val;//这种没有重复 */
23 }
24 ll lowbit(ll x){
25     return x&(-x);
26 }
27 ll qu_sum(ll x){
28     ll s=0;
29     while (x>0) {
30         s+=c[x];
31         x-=lowbit(x);
32     }
33     return s;
34 }
35 void updata(ll x){
36     while (x<=n) {
37         c[x]+=1;
38         x+=lowbit(x);
39     }
40 }
41 
42 ll ans;
43 int main () {
44     scanf("%lld",&n);
45     memset(c,0,sizeof(c));
46     for (int i=1;i<=n;i++) {
47         scanf("%lld",&num[i].val);
48         num[i].pos=i;
49     }
50     sort(num+1,num+n+1,cmp);
51     ans=0;
52     for (int i=1;i<=n;i++) {
53         updata(num[i].pos);
54         ans+=qu_sum(num[i].pos-1);
55     }
56     printf("%lld\n",ans);
57     return 0;
58 }

 有的时候,

n 太大,

可以用离散化来求,

下面是离散化例子,

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <string>
 4 #include <bits/stdc++.h>
 5 
 6 using namespace std;
 7 typedef long long ll;
 8 typedef unsigned long long ull;
 9 
10 #define MAXN 100005
11 ll c[MAXN],s[MAXN],a[MAXN],b[MAXN];
12 ll n;
13 int lowbit(int x) {return x&(-x);}
14 void updata(ll i,int x) {
15     while (i<=n) {
16         c[i]+=x;
17         i+=lowbit(i); 
18     }
19 }
20 ll qu_sum(ll x) {
21     ll sum=0;
22     while(x>0) {
23         sum+=c[x];
24         x-=lowbit(x);
25     }
26     return sum;
27 }
28 int main () {
29     //还是求逆序对 
30     scanf("%lld",&n);
31     ll ans=0;
32     for (int i=1;i<=n;i++) {
33         scanf("%lld",&a[i]);
34         a[i]+=1000000000;
35         b[i]=a[i];
36     }
37     sort(b+1,b+n+1);
38     int len=unique(b+1,b+n+1)-b-1; 
39     //STL库去重
40     for (int i=1;i<=n;i++) {
41         a[i]=lower_bound(b+1,b+len+1,a[i])-b;
42         updata(a[i],1);
43         ans+=(ll)i-qu_sum(a[i]);
44     }
45     printf("%lld\n",ans);
46     return 0;
47 }

树状数组还有好多扩展应用,

比如下面这个,

就是区间加,区间查询和,

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <string>
 4 #include <bits/stdc++.h>
 5 
 6 using namespace std;
 7 typedef long long ll;
 8 typedef unsigned long long ull;
 9 #define MAXN 100010
10 int a[MAXN],n,m;
11 ll c[2][MAXN],sum[MAXN];
12 ll lowbit(int x) {return x&(-x);}
13 ll asksum(int k,int x) {
14     ll ans=0;
15     for (;x;x-=lowbit(x)) {
16         ans+=c[k][x];
17     }
18     return ans;
19 }
20 void add(int k,int x,int y) {
21     for (;x<=n;x+=lowbit(x)){
22         c[k][x]+=y;
23     }
24 }
25 int main () {
26     scanf("%d%d",&n,&m);
27     for (int i=1;i<=n;i++) {
28         scanf("%d",&a[i]);
29         sum[i]=sum[i-1]+a[i];
30     }
31     for (int i=1,l,r,d;i<=m;i++){
32         char op[2];
33         cin>>op;
34         scanf("%d%d",&l,&r);
35         if (op[0]=='C') {
36             scanf("%d",&d);
37             //区间加 
38             add(0,l,d);
39             add(0,r+1,-d);
40             add(1,l,l*d);
41             add(1,r+1,-(r+1)*d);
42         }
43         else {
44             //区间求和
45             ll ans=0;
46             ans=sum[r]+(r+1)*asksum(0,r)-asksum(1,r);
47             ans-=sum[l-1]+l*asksum(0,l-1)-asksum(1,l-1);
48             printf("%lld\n",ans);
49         }
50     } 
51 
52     return 0;
53 }

这回就很全了,

 

最后还有归并排序求逆序对,

 

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <string>
 4 #include <bits/stdc++.h>
 5 
 6 using namespace std;
 7 typedef long long ll;
 8 typedef unsigned long long ull;
 9 #define MAXN 100001
10 ll ans,n;
11 int s[MAXN],v[MAXN];
12 //s[i]存数据,v[i]更改添加 
13 
14 void fenzhi(int x,int y) {
15     if(x==y) {
16         return;
17     }
18     int mid=(x+y)/2;
19     fenzhi(x,mid);
20     fenzhi(mid+1,y);
21     int p=x,j=x,k=mid+1;
22     while(j<=mid&&k<=y) {
23         if(s[j]>s[k]) {
24             ans+=mid-j+1;
25             v[p++]=s[k++];
26         } else v[p++]=s[j++];
27     }
28     while(j<=mid) {
29         v[p++]=s[j++];
30     }
31     while(k<=y) {
32         v[p++]=s[k++];
33     }
34     for(int i=x; i<=y; i++) {
35         s[i]=v[i];
36     }
37 }
38 int main() {
39     cin>>n;
40     for(int i=1; i<=n; i++) {
41         cin>>s[i];
42     }
43     fenzhi(1,n);
44     printf("%lld\n",ans);
45     return 0;
46 }

终于清了,

逆序对还要常看看

再来几个吧：

 

单点修改,区间查询

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
using namespace std;
int n,m;
int c[500005];
int lowbit(int x) {
    return (-x)&x;
}
void add(int pos,int x) {
    while(pos<=n) {
        c[pos]+=x;
        pos+=lowbit(pos);
    }
}
void input() {
    int x;
    for(int i=1; i<=n; i++) {
        scanf("%d",&x);
        add(i,x);
    }
}
int query(int pos) {
    int res=0;
    while(pos>0) {
        res+=c[pos];
        pos-=lowbit(pos);
    }
    return res;
}
int main() {
    scanf("%d%d",&n,&m);
    input();
    int f,x,y;
    for(int i=1; i<=m; i++) {
        scanf("%d%d%d",&f,&x,&y);
        if(f==1) add(x,y);
        else if(f==2) cout<<query(y)-query(x-1)<<endl;
    }
    return 0;
}

区间修改,单点查询：

#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;
int c[500005];
int n,m;
int lowbit(int x) {
    return x&(-x);
}
void add(int pos,int x) {
    while(pos<=n) {
        c[pos]+=x;
        pos+=lowbit(pos);
    }
}
int query(int pos) {
    int res=0;
    while(pos>0) {
        res+=c[pos];
        pos-=lowbit(pos);
    }
    return res;
}
int main() {
    scanf("%d%d",&n,&m);
    int x=0,y;
    for(int i=1; i<=n; i++) {
        scanf("%d",&y);
        add(i,y-x);
        x=y;
    }
    int opt,k;
    for(int i=1; i<=m; i++) {
        scanf("%d",&opt);
        if(opt==1) {
            scanf("%d%d%d",&x,&y,&k);
            add(x,k);
            add(y+1,-k);
        } else if(opt==2) {
            scanf("%d",&x);
            printf("%d\n",query(x));
        }
    }
    return 0;
}

以及

 

区间修改,区间查询：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long long c[200005][2];
int n,q;
int lowbit(int x) {
    return x&(-x);
}
void add(int pos,int x,int f) {
    while(pos<=n) {
        c[pos][f]+=x;
        pos+=lowbit(pos);
    }
}
long long query(int pos,int f) {
    long long res=0;
    while(pos>0) {
        res+=c[pos][f];
        pos-=lowbit(pos);
    }
    return res;
}
long long ask(int pos) {
    long long res=(pos+1)*query(pos,0)-query(pos,1);
    return res;
}
int main() {
    scanf("%d",&n);
    int a=0,b,opt,x;
    for(int i=1; i<=n; i++) {
        scanf("%d",&b);
        add(i,b-a,0);
        add(i,(b-a)*i,1);
        a=b;
    }
    scanf("%d",&q);
    for(int i=1; i<=q; i++) {
        scanf("%d",&opt);
        if(opt==1) {
            scanf("%d%d%d",&a,&b,&x);
            add(a,x,0);
            add(b+1,-x,0);
            add(a,x*a,1);
            add(b+1,-x*(b+1),1);
        } else if(opt==2) {
            scanf("%d%d",&a,&b);
            printf("%lld\n",ask(b)-ask(a-1));
        }
    }
    return 0;
}

 

posted @ 2018-10-16 09:13  codemaker_li  阅读(157)  评论(0)    收藏  举报