bzoj1452

二维树状数组or树套树

树套树在bzoj上T了，在洛谷上AC了(卡着1000ms的上限跑的，交了n发才过)

二维树状数组跑的飞快

二维树状数组代码：

/**************************************************************

    Problem: 1452

    User: syh0313

    Language: C++

    Result: Accepted

    Time:4672 ms

    Memory:42960 kb

****************************************************************/

 

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <string>

#include <cmath>

using namespace std;

const int maxn=310;

int n,m,q,col[maxn][maxn],a[110][maxn][maxn];

int lowbit(int x){return x&(-x);}

void add(int k,int x,int y,int kk)

{

    while (x<=n)

    {

        for (int i=y;i<=m;i+=lowbit(i)) a[k][x][i]+=kk;

        x+=lowbit(x);

    }

}

int qury(int kk,int x,int y)

{

    int sum=0;

    while (x)

    {

        for (int i=y;i;i-=lowbit(i)) sum+=a[kk][x][i];

        x-=lowbit(x);

    }

return sum;

}

int main()

{

    scanf("%d%d",&n,&m);

    for (int i=1;i<=n;i++)

     for  (int j=1;j<=m;j++)

     {

        scanf("%d",&col[i][j]);

        add(col[i][j],i,j,1);

     }

    scanf("%d",&q);

    while (q--)

    {

        int ff; scanf("%d",&ff);

        if (ff==1)

        {

            int xx,yy,kk; scanf("%d%d%d",&xx,&yy,&kk);

            add(col[xx][yy],xx,yy,-1);

            add(kk,xx,yy,1); col[xx][yy]=kk;

        }

        else

        {

            int x,xx,y,yy,kk; scanf("%d%d%d%d%d",&x,&xx,&y,&yy,&kk);

            printf("%d\n",qury(kk,xx,yy)-qury(kk,x-1,yy)-qury(kk,xx,y-1)+qury(kk,x-1,y-1));

        }

    }

return 0;

}

树套树洛谷AC代码：

/**************************************************************     Problem: 1452     User: syh0313     Language: C++     Result: Time_Limit_Exceed ****************************************************************/   #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #define lch a[n].lc #define rch a[n].rc using namespace std; const int maxn=310; struct da{int lc,rc,sum;}a[100000*maxn]; struct daa{int lc,rc,sum;}b[10000*maxn]; int n,m,col[1100],root[10000*maxn],topt,cnt,q,co[maxn][maxn]; inline void updata(int n){a[n].sum=a[lch].sum+a[rch].sum;} inline void tree_add_2(int &n,int l,int r,int lc,int k) {     if (!n) n=++cnt;     if (l==r) {a[n].sum+=k; return;}     int mid=(l+r)>>1;     if (lc<=mid) tree_add_2(lch,l,mid,lc,k);      else tree_add_2(rch,mid+1,r,lc,k);     updata(n); } inline void tree_add(int &n,int l,int r,int lx,int ly,int k) {     if (!n) n=++topt;     tree_add_2(root[n],1,m,ly,k);     if (l==r) return;     int mid=(l+r)>>1;     if (lx<=mid) tree_add(b[n].lc,l,mid,lx,ly,k);      else tree_add(b[n].rc,mid+1,r,lx,ly,k); } inline int qury_2(int &n,int L,int R,int l,int r) {     if (!n) return 0;     if (L==l && R==r) return a[n].sum;     int mid=(L+R)>>1;     if (r<=mid) return qury_2(lch,L,mid,l,r);      else if (l>=mid+1) return qury_2(rch,mid+1,R,l,r); return qury_2(lch,L,mid,l,mid)+qury_2(rch,mid+1,R,mid+1,r); } inline int qury(int n,int L,int R,int l,int r,int y1,int y2) {     if (!n) return 0;     if (L==l && R==r) return qury_2(root[n],1,m,y1,y2);     int mid=(L+R)>>1;     if (r<=mid) return qury(b[n].lc,L,mid,l,r,y1,y2);      else if (l>=mid+1) return qury(b[n].rc,mid+1,R,l,r,y1,y2); return qury(b[n].lc,L,mid,l,mid,y1,y2)+qury(b[n].rc,mid+1,R,mid+1,r,y1,y2); } inline char nc(void) {     static char ch[100010],*p1=ch,*p2=ch;     return p1==p2&&(p2=(p1=ch)+fread(ch,1,100010,stdin),p1==p2)?EOF:*p1++; } inline int read() {     int xxx=0,ff=1; char c=nc();     while (c<'0' || c>'9') {if (c=='-') ff=-1; c=nc();}     while (c>='0' && c<='9') {xxx=(xxx<<1)+(xxx<<3)+c-'0'; c=nc();} return xxx*ff; } int main() {     n=read(); m=read(); register int i,j;     for (i=1;i<=n;i++)      for (j=1;j<=m;j++)      {         int xx; xx=read(); co[i][j]=xx;         tree_add(col[xx],1,n,i,j,1);      }     q=read();     while (q--)     {         int ff; ff=read();         if (ff==1)         {             int xx,yy,kk; xx=read(); yy=read(); kk=read();             tree_add(col[co[xx][yy]],1,n,xx,yy,-1);             tree_add(col[kk],1,n,xx,yy,1); co[xx][yy]=kk;         }         else         {             int x,y,xx,yy,kk;             x=read(); xx=read(); y=read(); yy=read(); kk=read();             printf("%d\n",qury(col[kk],1,n,x,xx,y,yy));         }     } return 0; }