BZOJ3295: [Cqoi2011]动态逆序对

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 100000+5
 26 
 27 #define maxm 4000000+5
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 44 
 45 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
 46 
 47 #define mod 1000000007
 48 
 49 using namespace std;
 50 
 51 inline int read()
 52 
 53 {
 54 
 55     int x=0,f=1;char ch=getchar();
 56 
 57     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 58 
 59     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 60 
 61     return x*f;
 62 
 63 }
 64 int n,m,tot,a[maxn],b[maxn],c[maxn],d[maxn],sa[maxn];
 65 ll ans[maxn];
 66 int s[maxm],rt[maxn],v[maxm],rnd[maxm],l[maxm],r[maxm];
 67 inline void add(int x){for(;x<=n;x+=x&(-x))d[x]++;}
 68 inline int sum(int x){int t=0;for(;x;x-=x&(-x))t+=d[x];return t;}
 69 inline void pushup(int k)
 70 {
 71     s[k]=s[l[k]]+s[r[k]]+1;
 72 }
 73 inline void lturn(int &k)
 74 {
 75     int t=r[k];r[k]=l[t];l[t]=k;pushup(k);pushup(t);k=t;
 76 }
 77 inline void rturn(int &k)
 78 {
 79     int t=l[k];l[k]=r[t];r[t]=k;pushup(k);pushup(t);k=t;
 80 }
 81 inline void insert(int &k,int x)
 82 {
 83     if(!k)
 84     {
 85         k=++tot;s[k]=1;v[k]=x;rnd[k]=rand();l[k]=r[k]=0;
 86         return;
 87     }
 88     s[k]++;
 89     if(x<v[k]){insert(l[k],x);if(rnd[k]<rnd[l[k]])rturn(k);}
 90     else {insert(r[k],x);if(rnd[k]<rnd[r[k]])lturn(k);}
 91 }
 92 inline int rank(int k,int x)
 93 {
 94     if(!k)return 0;
 95     if(x<v[k])return rank(l[k],x);
 96     else return s[l[k]]+1+rank(r[k],x);
 97 }
 98 inline void add(int x,int y)
 99 {
100     for(;x<=n;x+=x&(-x))insert(rt[x],y);
101 }
102 inline int query1(int x,int y)
103 {
104     int t=0;
105     for(;x;x-=x&(-x))t+=s[rt[x]]-rank(rt[x],y);
106     return t;
107 }
108 inline int query2(int x,int y)
109 {
110     int t=0;
111     for(;x;x-=x&(-x))t+=rank(rt[x],y);
112     return t;
113 }
114 
115 int main()
116 
117 {
118 
119     freopen("input.txt","r",stdin);
120 
121     freopen("output.txt","w",stdout);
122 
123     n=read();m=read()-1;
124     for1(i,n)a[i]=read(),sa[a[i]]=i;
125     for1(i,m)b[i]=sa[read()],c[b[i]]=1;
126     for1(i,n)if(!c[i])
127     {
128        ans[m+1]+=(ll)(sum(n)-sum(a[i]));
129        add(a[i]);
130        add(i,a[i]);
131     }
132     for3(i,m,1)
133     {
134         ans[i]+=(ll)query1(b[i],a[b[i]]);
135         add(b[i],a[b[i]]);
136     }
137     memset(rt,0,sizeof(rt));tot=0;
138     for1(i,n>>1)swap(a[i],a[n+1-i]);
139     for1(i,n)if(!c[n+1-i])add(i,a[i]);
140     for3(i,m,1)
141     {
142         b[i]=n+1-b[i];
143         ans[i]+=(ll)query2(b[i],a[b[i]]);
144         add(b[i],a[b[i]]);
145     }
146     for3(i,m,1)ans[i]+=ans[i+1];
147     for1(i,m+1)printf("%lld\n",ans[i]);
148 
149     return 0;
150 
151 }  

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 100000+5
 26 
 27 #define maxm 4000000+5
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 44 
 45 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
 46 
 47 #define mod 1000000007
 48 
 49 using namespace std;
 50 
 51 inline int read()
 52 
 53 {
 54 
 55     int x=0,f=1;char ch=getchar();
 56 
 57     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 58 
 59     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 60 
 61     return x*f;
 62 
 63 }
 64 int n,m,tot,a[maxn],b[maxn],c[maxn],d[maxn],sa[maxn];
 65 ll ans[maxn];
 66 int s[maxm],rt[maxn],v[maxm],rnd[maxm],l[maxm],r[maxm];
 67 inline void add(int x){for(;x<=n;x+=x&(-x))d[x]++;}
 68 inline int sum(int x){int t=0;for(;x;x-=x&(-x))t+=d[x];return t;}
 69 inline void pushup(int k)
 70 {
 71     s[k]=s[l[k]]+s[r[k]]+1;
 72 }
 73 inline void lturn(int &k)
 74 {
 75     int t=r[k];r[k]=l[t];l[t]=k;pushup(k);pushup(t);k=t;
 76 }
 77 inline void rturn(int &k)
 78 {
 79     int t=l[k];l[k]=r[t];r[t]=k;pushup(k);pushup(t);k=t;
 80 }
 81 inline void insert(int &k,int x)
 82 {
 83     if(!k)
 84     {
 85         k=++tot;s[k]=1;v[k]=x;rnd[k]=rand();l[k]=r[k]=0;
 86         return;
 87     }
 88     s[k]++;
 89     if(x<v[k]){insert(l[k],x);if(rnd[k]<rnd[l[k]])rturn(k);}
 90     else {insert(r[k],x);if(rnd[k]<rnd[r[k]])lturn(k);}
 91 }
 92 inline int rank(int k,int x)
 93 {
 94     if(!k)return 0;
 95     if(x<v[k])return rank(l[k],x);
 96     else return s[l[k]]+1+rank(r[k],x);
 97 }
 98 inline void add(int x,int y)
 99 {
100     for(;x<=n;x+=x&(-x))insert(rt[x],y);
101 }
102 inline int query1(int x,int y)
103 {
104     int t=0;
105     for(;x;x-=x&(-x))t+=s[rt[x]]-rank(rt[x],y);
106     return t;
107 }
108 inline int query2(int x,int y)
109 {
110     int t=0;
111     for(;x;x-=x&(-x))t+=rank(rt[x],y);
112     return t;
113 }
114 
115 int main()
116 
117 {
118 
119     freopen("input.txt","r",stdin);
120 
121     freopen("output.txt","w",stdout);
122 
123     n=read();m=read()-1;
124     for1(i,n)a[i]=read(),sa[a[i]]=i;
125     for1(i,m)b[i]=sa[read()],c[b[i]]=1;
126     for1(i,n)if(!c[i])
127     {
128        ans[m+1]+=(ll)(sum(n)-sum(a[i]));
129        add(a[i]);
130        add(i,a[i]);
131     }
132     for3(i,m,1)
133     {
134         ans[i]+=(ll)(query1(b[i],a[b[i]])+query2(n,a[b[i]])-query2(b[i],a[b[i]]));
135         add(b[i],a[b[i]]);
136     }
137     for3(i,m,1)ans[i]+=ans[i+1];
138     for1(i,m+1)printf("%lld\n",ans[i]);
139 
140     return 0;
141 
142 }