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 ",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 ",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 ",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 ",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 ",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 ",ask(b)-ask(a-1)); } } return 0; }