Problem--ZOJ1610 Count the Colors(Segment Tree）

Problem--ZOJ1610 Count the Colors(Segment Tree）

ZOJ1610--Count the Colors
Analysis:
This is obviously a problem that we can solve using the data structure called Segment Tree.Specifically for this problem,we can build a Segment Tree according to the input datas.First,we get all datas from the input,and find out the longest segment modified.Then we build the tree with the longest segment as its root.After that,we can implement each manipulation one by one,using the Insert_ Segment function.Note that every single modification should be carried out exactly from the root of the tree,due to the charateristics of the Segment Tree.Finally,we use a function Count_Color to calculate the number of segments on which each color is visible.Note that we should merge two joint segments into one when counting.This process is important and necessary only because we did not do such thing when 'painting' the segments,i.e,when a segment [a,b](a<b) is painted with a single color,we did not paint all its successive segments with the same color,which we could've done.However,for this problem solely,we really should not do that when painting,due to the fact that this problem requires only one 'query'.That being true,why not save the time and do that job only once?
Code (in C++) :

  1  #include < iostream >
  2  #include < cstring >
  3  #define  MAXSIZE 32000
  4  using   namespace  std;
  5  struct  segment
  6  {
  7       int  color; // color>=0 indicates that this segment is monochrome,while color==-1 indicates polychrome.A value -2 indicates the original state
  8       int  start,end; // the starting and ending point of a segment
  9  }seg[MAXSIZE];
 10  struct  Dye
 11  {
 12       int  left,right,chrome;
 13  }manipulation[ 8000 ],visible[ 8000 ];
 14  void  Build( int  num, int  left, int  right)
 15  {
 16      seg[num].start = left;seg[num].end = right;seg[num].color =- 2 ; // set the original state of the current segment
 17       if (left + 1 == right)  return ; // if the segment cannot be divided any more,return
 18       int  mid = (left + right) / 2 ; // divide it into two
 19      Build(num * 2 ,left,mid);Build(num * 2 + 1 ,mid,right); // Build the left and right child of the current segment,respectively
 20  }
 21  /* This function is used to build a segment tree,with its parameters representing the index of
 22  the root(num),the starting point of the total segment(left),and the ending point of the total
 23  segment(right),respectively */
 24  void  Insert_Segment( int  num, int  left, int  right, int  chrome)
 25  {
 26       if (seg[num].color != chrome)
 27      {
 28           if (left == seg[num].start  &&  right == seg[num].end)
 29          {
 30              seg[num].color = chrome;
 31               if (num == 1 )  return ;
 32               if (num % 2 == 0 )
 33              {
 34                   if (seg[num + 1 ].color == chrome) seg[num / 2 ].color = chrome;
 35              }
 36               else   if (seg[num - 1 ].color == chrome) seg[num / 2 ].color = chrome;
 37               return ;
 38          } // if the color segment is exactly the current segment,then color it and return
 39           else   if (seg[num].color >= 0 )
 40          {
 41              seg[ 2 * num].color = seg[num].color;
 42              seg[ 2 * num + 1 ].color = seg[num].color;
 43          } // if the color segment does not match the current segment perfectly,and the current segment was monochrome
 44            // then paint its children its original color,and move on
 45          seg[num].color =- 1 ; // two scenarios will lead us here:the segment has its original color
 46                             // and gets mixed by the current painting,or the segment was a polychrome
 47                             // one.Either way,we should mark it as polychrome here and move on
 48           int  mid = (seg[num].start + seg[num].end) / 2 ; // find the middle point of the segment,looking to paint it
 49           if (mid >= right)
 50          {
 51              Insert_Segment( 2 * num,left,right,chrome);
 52               if (seg[ 2 * num].color == seg[ 2 * num + 1 ].color  &&  seg[ 2 * num].color >= 0 )
 53                  seg[num].color = seg[ 2 * num].color;
 54          }
 55           else   if (mid <= left)
 56          {
 57              Insert_Segment( 2 * num + 1 ,left,right,chrome);
 58               if (seg[ 2 * num].color == seg[ 2 * num + 1 ].color  &&  seg[ 2 * num].color >= 0 )
 59                  seg[num].color = seg[ 2 * num].color;
 60          }
 61           else
 62          {
 63              Insert_Segment( 2 * num,left,mid,chrome);
 64              Insert_Segment( 2 * num + 1 ,mid,right,chrome);
 65               if (seg[ 2 * num].color == seg[ 2 * num + 1 ].color  &&  seg[ 2 * num].color >= 0 )
 66                  seg[num].color = seg[ 2 * num].color; // if the two child segments end up with the same
 67                                                   // monochrome,update the father segment
 68          } // decide in which area should the paint be carried on
 69      } // all the above should only be done when the chrome to be painted is not the same as the
 70        // original color of the current segment
 71  }
 72  /* This function is used to insert paint area into the segment tree we have built,with its parameters
 73  representing the index of the root of the tree(num),the starting and ending point of the paint
 74  segment(left,right),as well as the color of the paint(chrome),respectively */
 75  void  Count_Color( int  num, int  left, int  right)
 76  {
 77       if (left + 1 == right  &&  seg[num].color ==- 2 )  return ;
 78       if (seg[num].color >= 0 )
 79      {
 80           // printf("%d %d %d\n",seg[num].start,seg[num].end,seg[num].color);
 81           if (seg[num].start == visible[seg[num].color].right)
 82              visible[seg[num].color].right = seg[num].end;
 83           else   if (seg[num].end == visible[seg[num].color].left)
 84              visible[seg[num].color].left = seg[num].start;
 85           else
 86          {
 87              visible[seg[num].color].chrome ++ ;
 88              visible[seg[num].color].left = seg[num].start;
 89              visible[seg[num].color].right = seg[num].end;
 90          }
 91           return ;
 92      } // if the current segment if monochrome,then the color of it is visible,obviously.In addition,
 93        // all the segments who are the children of it stays invisible.Consequencely,the counting
 94        // process should be ended.
 95       else
 96      {
 97           int  mid = (left + right) / 2 ;
 98          Count_Color(num * 2 ,left,mid);
 99          Count_Color(num * 2 + 1 ,mid,right);
100      }
101       // Either the current segment is polychrome,or remaining untouched,the counting should process
102       // down to its children.
103  }
104  /* This function is used to decide on the visibility together with the number of segments of each
105  color,with its parameters representing the index of the root of the tree(num),the startpoint and
106  the endpoint of the total segment(left,right),respectively. */
107  void  Output( int  left, int  right)
108  {
109       for ( int  i = left;i <= right;i ++ )
110      {
111           if (visible[i].chrome)
112          {
113              printf( " %d %d\n " ,i,visible[i].chrome);
114          }
115      }
116      printf( " \n " );
117  }
118  int  main()
119  {
120       int  n,mleft,mright,lcolor,rcolor;
121       while (scanf( " %d " , & n) == 1 )
122      {
123          mleft = lcolor = 8001 ;mright = rcolor = 0 ;
124           for ( int  i = 0 ;i < n;i ++ )
125          {
126              scanf( " %d %d %d " , & manipulation[i].left, & manipulation[i].right, & manipulation[i].chrome);
127               if (mleft > manipulation[i].left) mleft = manipulation[i].left;
128               if (mright < manipulation[i].right) mright = manipulation[i].right;
129               if (lcolor > manipulation[i].chrome) lcolor = manipulation[i].chrome;
130               if (rcolor < manipulation[i].chrome) rcolor = manipulation[i].chrome;
131          } // Input all the manipulations,and find out the leftmost and rightmost endpoint of the area
132            // together with the lowest and highest index of the color used
133          Build( 1 ,mleft,mright); // Build the segment tree due to the largest segment area
134           for ( int  i = 0 ;i < n;i ++ )
135          {
136              Insert_Segment( 1 ,manipulation[i].left,manipulation[i].right,manipulation[i].chrome);
137          } // Implement the manipulations
138           for ( int  i = lcolor;i <= rcolor;i ++ )
139          {
140              visible[i].chrome = 0 ;
141              visible[i].left = visible[i].right =- 1 ;
142          }
143          Count_Color( 1 ,mleft,mright); // Count the colors visible
144          Output(lcolor,rcolor); // Output the results
145      }
146       return   0 ;
147  }
148