• POJ2777 Count Color[线段树求整段区间]


    Count Color
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 27655   Accepted: 8263

    Description

    Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem.

    There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board:

    1. "C A B C" Color the board from segment A to segment B with color C.
    2. "P A B" Output the number of different colors painted between segment A and segment B (including).

    In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your.

    Input

    First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.

    Output

    Ouput results of the output operation in order, each line contains a number.

    Sample Input

    2 2 4
    C 1 1 2
    P 1 2
    C 2 2 2
    P 1 2
    

    Sample Output

    2
    1
    

    Source

    题意:整段区间染色

    code:

      1 #include<iostream>
      2 using namespace std;
      3 
      4 int l,t,o;
      5 
      6 struct Node
      7 {
      8     int l,r;
      9     int col;
     10 };
     11 struct Node node[100010*4];
     12 
     13 void build(int left,int right,int cur=1)
     14 {
     15     node[cur].l=left;
     16     node[cur].r=right;
     17     node[cur].col=0;
     18     if(left==right)
     19         return;
     20     int mid=(left+right)/2;
     21     build(left,mid,cur*2);
     22     build(mid+1,right,cur*2+1);
     23 }
     24 
     25 void update(int left,int right,int col,int cur=1)
     26 {
     27     if(left<=node[cur].l&&right>=node[cur].r)
     28     {
     29         node[cur].col=col;
     30         return;
     31     }
     32     if(node[cur].col>0)
     33     {
     34         node[cur*2].col=node[cur*2+1].col=node[cur].col;
     35         node[cur].col=0;
     36     }
     37     int mid=(node[cur].l+node[cur].r)/2;
     38     if(left>mid)
     39         update(left,right,col,cur*2+1);
     40     else if(right<=mid)
     41         update(left,right,col,cur*2);
     42     else
     43     {
     44         update(left,mid,col,cur*2);
     45         update(mid+1,right,col,cur*2+1);
     46     }
     47 }
     48 
     49 bool vst[40];
     50 int query(int left,int right,int cur=1)
     51 {
     52     int sum=0;
     53     if(node[cur].col)
     54     {
     55         if(!vst[node[cur].col])
     56         {
     57             sum++;
     58             vst[node[cur].col]=true;
     59         }
     60         return sum;
     61     }
     62     int mid=(node[cur].l+node[cur].r)/2;
     63     if(left>mid)
     64         sum+=query(left,right,cur*2+1);
     65     else if(right<=mid)
     66         sum+=query(left,right,cur*2);
     67     else
     68     {
     69         sum+=query(left,right,cur*2);
     70         sum+=query(left,right,cur*2+1);
     71     }
     72     return sum;
     73 }
     74 
     75 int main()
     76 {
     77     scanf("%d%d%d",&l,&t,&o);
     78     build(1,l);
     79     node[1].col=1;
     80     while(o--)
     81     {
     82         int a,b;
     83         char ch;
     84         getchar();
     85         scanf("%c%d%d",&ch,&a,&b);
     86         if(a>b)
     87         {
     88             int temp=a;
     89             a=b;
     90             b=temp;
     91         }
     92         if(ch=='C')
     93         {
     94             int c;
     95             scanf("%d",&c);
     96             update(a,b,c);
     97         }
     98         else
     99         {
    100             memset(vst,false,sizeof(vst));
    101             printf("%d\n",query(a,b));
    102         }
    103     }
    104     return 0;
    105 }
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  • 原文地址:https://www.cnblogs.com/XBWer/p/2665149.html
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