bzoj3939: [Usaco2015 Feb]Cow Hopscotch

题目

1.cdq分治

每次分成左右两块，用左边那块更新右边的

#include<bits/stdc++.h>
using namespace std;
const int M=1e9+7;
int f[1002][1002],sum[1002][1002],cnt[1000002],i,j,n,m,k,a[1002][1002];
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
    int x=0,fl=1;char ch=gc();
    for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
    for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
    return x*fl;
}
void cdq(int l,int r){
    if (l==r) return;
    int mid=l+r>>1;
    cdq(l,mid);
    for (int i=1;i<=n;i++)
        for (int j=l;j<=r;j++) cnt[a[i][j]]=0;
    int sum=0;
    for (int i=2;i<=n;i++){
        for (int j=l;j<=mid;j++) (sum+=f[i-1][j])%=M,(cnt[a[i-1][j]]+=f[i-1][j])%=M;
        for (int j=mid+1;j<=r;j++) (f[i][j]+=(sum-cnt[a[i][j]])%M)%=M;
    }
    cdq(mid+1,r);
}
int main(){
    n=rd();m=rd();k=rd();
    for (i=1;i<=n;i++)
        for (j=1;j<=m;j++) a[i][j]=rd();
    f[1][1]=1;
    cdq(1,m);
    printf("%d",(f[n][m]+M)%M);
}

2.线段树

动态开点

#include<bits/stdc++.h>
using namespace std;
const int M=1e9+7;
int f[1002][1002],sum[10000002],son[10000002][2],tot,i,j,n,m,k,a[1002][1002],rt[1000002];
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
    int x=0,fl=1;char ch=gc();
    for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
    for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
    return x*fl;
}
inline int add(int x,int y){x+=y;if(x>=M)x-=M;return x;}
int query(int t,int x,int y,int l,int r){
    if (x>y || l>r || !t) return 0;
    if (x==l && y==r) return sum[t];
    int mid=l+r>>1;
    if (y<=mid) return query(son[t][0],x,y,l,mid);
    if (mid<x) return query(son[t][1],x,y,mid+1,r);
    return (query(son[t][0],x,mid,l,mid)+query(son[t][1],mid+1,y,mid+1,r))%M;
}
void update(int &t,int x,int l,int r,int v){
    if (!t) t=++tot;
    sum[t]=add(sum[t],v);
    if (l==r) return;
    int mid=l+r>>1;
    if (x<=mid) update(son[t][0],x,l,mid,v);
    else update(son[t][1],x,mid+1,r,v);
}
int main(){
    n=rd();m=rd();k=rd();
    for (i=1;i<=n;i++)
        for (j=1;j<=m;j++) a[i][j]=rd();
    f[1][1]=1;
    update(rt[0],1,1,m,1);
    update(rt[a[1][1]],1,1,m,1);
    for (i=2;i<=n;i++){
        for (j=1;j<=m;j++) f[i][j]=add(query(rt[0],1,j-1,1,m)-query(rt[a[i][j]],1,j-1,1,m),M);
        for (j=1;j<=m;j++) update(rt[0],j,1,m,f[i][j]),update(rt[a[i][j]],j,1,m,f[i][j]);
    }
    printf("%d",f[n][m]);
}