POJ 1195 2维线段树(树套树实现) 树状数组

   1:  #include <stdio.h>

   2:  #include <string.h>

   3:  #include <stdlib.h>

   4:  #include <algorithm>

   5:  #include <iostream>

   6:  using namespace std;

   7:   

   8:  #define LL(a)   a<<1

   9:  #define RR(a)   a<<1|1

  10:  const int MaxL = 1005;

  11:  struct sub_Seg_tree_node

  12:  {

  13:      int subleft,subright; // ydown = subleft, yup = subright

  14:  //    int maxval;  // the maval of the rect x [left,right], y [subleft, subright]

  15:      int sum;     // the sum of the num of point in rect x [left, right], y [subleft, subright]

  16:  };

  17:   

  18:  struct Seg_tree_node

  19:  {

  20:      int left,right;       // left = xleft, right = xright

  21:      sub_Seg_tree_node T[MaxL<<2];

  22:  } TT[MaxL<<2];

  23:   

  24:  void sub_build(int subl, int subr, int subidx, int idx)

  25:  {

  26:      TT[idx].T[subidx].subleft=subl;

  27:      TT[idx].T[subidx].subright=subr;

  28:  //    TT[idx].T[subidx].maxval=-1;

  29:      TT[idx].T[subidx].sum=0;

  30:      if(subl==subr)  return;

  31:      int mid=(subl+subr)>>1;

  32:      sub_build(subl, mid, LL(subidx), idx);

  33:      sub_build(mid+1, subr, RR(subidx), idx);

  34:  }

  35:   

  36:  void build(int l, int r, int subl, int subr, int idx)

  37:  {

  38:      TT[idx].left = l, TT[idx].right = r;

  39:      sub_build(subl, subr, 1, idx);

  40:      if(l==r) return;

  41:      int mid=(l+r)>>1;

  42:      build(l,mid,subl,subr, LL(idx));

  43:      build(mid+1,r,subl,subr, RR(idx));

  44:  }

  45:   

  46:  void sub_update(int y, int val, int subidx, int idx)

  47:  {

  48:      TT[idx].T[subidx].sum += val;

  49:  //    TT[idx].T[subidx].maxval=max(TT[idx].T[subidx].maxval, val);

  50:      if(TT[idx].T[subidx].subleft == TT[idx].T[subidx].subright) return;

  51:      int mid=(TT[idx].T[subidx].subleft+TT[idx].T[subidx].subright)>>1;

  52:      if(y <= mid)

  53:          sub_update(y, val, LL(subidx), idx);

  54:      else

  55:          sub_update(y, val, RR(subidx), idx);

  56:  }

  57:   

  58:  void update(int x, int y, int val, int idx)

  59:  {

  60:      sub_update(y, val, 1, idx);

  61:      if(TT[idx].left == TT[idx].right) return;

  62:      int mid=(TT[idx].left+TT[idx].right)>>1;

  63:      if( x<=mid)

  64:          update(x,y,val, LL(idx) );

  65:      else

  66:          update(x,y, val,RR(idx));

  67:  }

  68:   

  69:  //int sub_query(int subl, int subr, int subidx, int idx)

  70:  //{

  71:  //    if(TT[idx].T[subidx].subleft == subl && TT[idx].T[subidx].subright == subr)

  72:  //        return TT[idx].T[subidx].maxval;

  73:  //    int mid = (TT[idx].T[subidx].subleft + TT[idx].T[subidx].subright)>>1;

  74:  //    if(subr <= mid)

  75:  //        return sub_query(subl, subr, LL(subidx), idx);

  76:  //    else if(subl > mid)

  77:  //        return sub_query(subl, subr, RR(subidx), idx);

  78:  //    else

  79:  //        return max(sub_query(subl, mid, LL(subidx), idx),sub_query(mid+1, subr, RR(subidx), idx));

  80:  //}

  81:  //

  82:  //int query(int l, int r, int subl, int subr, int idx)

  83:  //{

  84:  //    if(TT[idx].left == l && TT[idx].right == r)

  85:  //        return sub_query(subl, subr, 1, idx);

  86:  //    int mid = (TT[idx].left + TT[idx].right)>>1;

  87:  //    if(r <= mid)

  88:  //        return query(l,r, subl,subr, LL(idx));

  89:  //    else if(l > mid)

  90:  //        return query(l, r, subl, subr, RR(idx));

  91:  //    else

  92:  //        return max(query(l, mid, subl, subr, LL(idx)),query(mid+1, r, subl,subr, RR(idx)));

  93:  //}

  94:   

  95:  int sub_query(int subl, int subr, int subidx, int idx)

  96:  {

  97:      if(TT[idx].T[subidx].subleft == subl && TT[idx].T[subidx].subright == subr)

  98:          return TT[idx].T[subidx].sum;

  99:      int mid = (TT[idx].T[subidx].subleft + TT[idx].T[subidx].subright)>>1;

 100:      if(subr <= mid)

 101:          return sub_query(subl, subr, LL(subidx), idx);

 102:      else if(subl > mid)

 103:          return sub_query(subl, subr, RR(subidx), idx);

 104:      else

 105:          return sub_query(subl, mid, LL(subidx), idx) + sub_query(mid+1, subr, RR(subidx), idx);

 106:  }

 107:   

 108:  int query(int l, int r, int subl, int subr, int idx)

 109:  {

 110:      if(TT[idx].left == l && TT[idx].right == r)

 111:          return sub_query(subl, subr, 1, idx);

 112:      int mid = (TT[idx].left + TT[idx].right)>>1;

 113:      if(r <= mid)

 114:          return query(l,r, subl,subr, LL(idx));

 115:      else if(l > mid)

 116:          return query(l, r, subl, subr, RR(idx));

 117:      else

 118:          return query(l, mid, subl, subr, LL(idx))+ query(mid+1, r, subl,subr, RR(idx));

 119:  }

 120:   

 121:  int x1[10005];

 122:  int y1[10005];

 123:  int x2[10005];

 124:  int y2[10005];

 125:  int px[10005];

 126:  int py[10005];

 127:  int pval[10005];

 128:  //

 129:  //int main()

 130:  //{

 131:  //    freopen("1.txt","r",stdin);

 132:  //    int NN,MM;

 133:  //    cin>>NN>>MM;

 134:  //

 135:  //    for(int i=0; i<NN; i++)

 136:  //        scanf("%d%d%d", &px[i],&py[i], &pval[i]);

 137:  //

 138:  //    for(int i=0; i<MM; i++)

 139:  //    {

 140:  //        scanf("%d%d%d%d", &x1[i], &y1[i], &x2[i], &y2[i]);

 141:  //        if(x1[i] > x2[i]) swap(x1[i], x2[i]);

 142:  //        if(y1[i] > y2[i]) swap(y1[i], y2[i]);

 143:  //    }

 144:  //

 145:  //    build(0, 1060, 0, 1060, 1);

 146:  //    for(int i=0; i<NN; i++)

 147:  //    {

 148:  //        cout<<"update point "<<px[i]<<" "<<py[i]<<endl;

 149:  //        update(px[i],py[i],pval[i], 1);

 150:  //    }

 151:  //

 152:  //    for(int i=0; i<MM; i++)

 153:  //    {

 154:  //        cout<<"query ret "<<x1[i]<<" "<<x2[i]<<" "<<y1[i]<<" "<<y2[i]<<endl;

 155:  //        cout<<"Ret "<<query(x1[i],x2[i],y1[i],y2[i],1)<<endl;

 156:  //    }

 157:  //

 158:  //    return 0;

 159:  //}

 160:  int s;

 161:  int main()

 162:  {

 163:  //    freopen("1.txt","r",stdin);

 164:      int i;

 165:      int sb;

 166:      int sb1,sb2,sb3,sb4;

 167:      scanf("%d",&sb);

 168:      while(sb!=3)

 169:      {

 170:          switch(sb)

 171:          {

 172:          case 0:

 173:              scanf("%d",&s);

 174:              build(0,s,0,s,1);

 175:              break;

 176:          case 1:

 177:              scanf("%d%d%d",&sb1,&sb2,&sb3);

 178:              update(sb1,sb2,sb3,1);

 179:              break;

 180:          case 2:

 181:              scanf("%d%d%d%d",&sb1,&sb2,&sb3,&sb4);

 182:              printf("%d\n",query(sb1,sb3,sb2,sb4,1));

 183:              break;

 184:          default:

 185:              goto ed;

 186:          }

 187:          scanf("%d",&sb);

 188:      }

 189:  ed:

 190:      ;

 191:      return 0;

 192:  }

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树状数组来做这个题目相对来说简单得多!  树状数组则清楚得多.相对来说二维线段树用得比较少,而且功能都比较简单,查询平面矩形包含几个点,查询矩形最大值之类的.

   1:  #include <iostream>

   2:  #include <stdio.h>

   3:  #include <string.h>

   4:  using namespace std;

   5:   

   6:  #define MAX 1026

   7:  int s[MAX][MAX];

   8:  int lowbit(int x)

   9:  {

  10:      return x&(x^(x-1));

  11:  }

  12:   

  13:  int N;

  14:   

  15:  void change(int a,int b,int delta)

  16:  {

  17:      for(int x=a; x<=N; x+=lowbit(x))

  18:          for(int y=b; y<=N; y+=lowbit(y))

  19:              s[x][y]+=delta;

  20:  }

  21:  int sum(int a,int b)

  22:  {

  23:      int res=0;

  24:      for(int x=a; x>0; x-=lowbit(x))

  25:          for(int y=b; y>0; y-=lowbit(y))

  26:              res+=s[x][y];

  27:      return res;

  28:  }

  29:  int main()

  30:  {

  31:      int n;

  32:      while(cin>>n)

  33:      {

  34:          if(n==0)

  35:          {

  36:              cin>>N; N++;

  37:              memset(s,0,sizeof(s));

  38:          }

  39:          else if(n==1)

  40:          {

  41:              int a,b,c;

  42:              cin>>a>>b>>c; a++; b++;

  43:              change(a,b,c);

  44:          }

  45:          else if(n==2)

  46:          {

  47:              int a,b,c,d;

  48:              cin>>a>>b>>c>>d;

  49:              a++;b++;c++;d++;

  50:              cout<<sum(c,d) + sum(a-1,b-1) - sum(a-1,d) - sum(c,b-1)<<endl;

  51:          }

  52:          else break;;

  53:      }

  54:      return 0;

  55:  }

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