bzoj3939: [Usaco2015 Feb]Cow Hopscotch
阿新 • • 發佈:2018-12-13
1.cdq分治
每次分成左右兩塊,用左邊那塊更新右邊的
#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.線段樹
動態開點
#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]);
}