POJ2777 Count Color（线段树）

题意：有一个长版，有两种操作，一种是在区间[a,b]上颜色，另外一种是询问区间有多少种颜色

思路：线段树维护一个color信息,其中color>0表示节点所指的区间都有某种颜色color, color=-1 表示该节点的子节点的颜色不一致.

#include<cstdio>#include<iostream>#include<cstdio>#include<iostream>#include<cstring>using namespace std;#define lson i<<1,l,m#define rson (i<<1)|1,m+1,rconst int maxn = 100000+100;bool vis[35];int cnt;int color[maxn<<2];void build(int i,int l,int r){    color[i]=1;    if (l==r)        return ;    int m = (l+r)>>1;    build(lson);    build(rson);}void pushdown(int i){    if (color[i]>0)        color[i<<1]=color[(i<<1)|1]=color[i];}void pushup(int i){    if (color[i<<1]==-1 || color[(i<<1)|1]==-1)        color[i]=-1;    else if (color[i<<1]==color[(i<<1)|1])        color[i]=color[i<<1];    else        color[i]=-1;}void update(int ql,int qr,int v,int i,int l,int r){    if (ql<=l && qr>=r)    {        color[i]=v;        return;    }    pushdown(i);    int m = (l+r)>>1;    if (ql<=m)        update(ql,qr,v,lson);    if (m<qr)        update(ql,qr,v,rson);    pushup(i);}void query(int ql,int qr,int i,int l,int r){    if (color[i]>0)    {        if (vis[color[i]]==false)            cnt++;        vis[color[i]]=true;        return;    }    int m = (l+r)>>1;    if (ql<=m)        query(ql,qr,lson);    if (qr>m)        query(ql,qr,rson);}int main(){    int n,t,q;    while (scanf("%d%d%d",&n,&t,&q)!=EOF)    {        build(1,1,n);        while (q--)        {            char str[10];            scanf("%s",str);            if (str[0]=='C')            {                int a,b,c;                scanf("%d%d%d",&a,&b,&c);                update(a,b,c,1,1,n);            }            else if (str[0]=='P')            {                int a,b;                scanf("%d%d",&a,&b);                memset(vis,0,sizeof(vis));                cnt = 0;                query(a,b,1,1,n);                printf("%d\n",cnt);            }        }    }}

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