POJ 2777 Count Color（线段树）

Count Color

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 47124 Accepted: 14246

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

Sample Output

21

Source

POJ Monthly--2006.03.26,dodo

题意：有一个长板子，多次操作，有两种操作，第一种是给从a到b那段染一种颜色c，另一种是询问a到b有多少种不同的颜色。

思路：线覆盖型线段树，不能传递。

代码：

#include<iostream>#include <string.h>using namespace std; struct edge{    int l,r;    int color;}tree[800001];int n,m,t;int v[31]; void build(int s,int t,int p){    tree[p].l=s;    tree[p].r=t;     tree[p].color=1;    if(t>s)    {        int mid=(s+t)>>1;        build(s,mid,p*2);        build(mid+1,t,p*2+1);    }} void update(int l,int r,int color,int p){    if(l==tree[p].l&&r==tree[p].r)    {        tree[p].color=color;                 return;    }    else    {        if(tree[p].color>0)        {            tree[2*p].color=tree[p].color;            tree[2*p+1].color=tree[p].color;            tree[p].color=0;        }        int mid=(tree[p].l+tree[p].r)>>1;        if(r<=mid)     update(l,r,color,2*p);        else        {            if(l>mid)     update(l,r,color,2*p+1);            else            {                update(l,mid,color,2*p);                update(mid+1,r,color,2*p+1);            }        }    }} void query(int l,int r,int p){    int i,j,k;    if(tree[p].color>0)    {        v[tree[p].color]=1;        return;    }    int mid=(tree[p].l+tree[p].r)>>1;    if(r<=mid)     query(l,r,2*p);    else        if(l>mid)     query(l,r,2*p+1);        else        {            query(l,mid,2*p);            query(mid+1,r,2*p+1);        }} int main(){    int i,j,k,s,tmp;    scanf("%d%d%d",&n,&t,&m);    build(1,n,1);    char c;    for(i=1;i<=m;i++)    {        cin>>c;        if(c=='C')        {            scanf("%d%d%d",&j,&k,&s);            if(j>k) swap(j,k);update(j,k,s,1);        }        else        {            scanf("%d%d",&j,&k);            if(j>k) swap(j,k);            memset(v,0,sizeof(v));            query(j,k,1);            j=0;            for(k=1;k<=t;k++)                if(v[k]) j++;            cout<<j<<endl;        }    }    return 0;}