poj2777

poj2777
区间修改,区间查询
mark=-1表示这个区间有多种颜色,否则就1种

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define For(i,a,b) for(register long long i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()
#define N 100010
using namespace std;
long long a[N],n,m,k,x,y,t,cnt;
bool vis[50];
char flag;
struct seg{
    long long mark;
}s[N<<3];

void in(long long &x){
    long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

inline void build(long long rt,long long l,long long r){
    s[rt].mark=1;
    if(l==r) return;
    long long mid=(l+r)>>1;
    build(rt<<1, l, mid);
    build(rt<<1|1, mid+1, r);
}

inline void pushdown(long long rt,long long l,long long r){
    if(s[rt].mark!=-1){
        s[rt<<1].mark=s[rt<<1|1].mark=s[rt].mark;
        s[rt].mark=-1;
    }
}

inline void update(long long rt,long long nl,long long nr,long long l,long long r,long long k){
    if(nr<l||nl>r)return;
    if(nl>=l&&nr<=r){
        s[rt].mark=k;
        return;
    }
    pushdown(rt,nl,nr);
    long long mid=(nl+nr)>>1;
    update(rt<<1, nl, mid, l, r, k);
    update(rt<<1|1, mid+1, nr, l, r, k);
}

inline void query(long long rt,long long nl,long long nr,long long l,long long r){
    if(nl>r||nr<l) return;
    if(s[rt].mark!=-1){
        vis[s[rt].mark]=1;
        return;
    }    
    long long mid=(nl+nr)>>1;
    query(rt<<1, nl, mid, l, r);
    query(rt<<1|1, mid+1, nr, l, r);
}

int main(){
    in(n);in(t);in(m);
    build(1,1,n);
    For(i,1,m){
        cin>>flag;
        if(flag=='C'){
            in(x);in(y);in(k);
            update(1,1,n,x,y,k);
        }
        else{
            memset(vis,0,sizeof(vis));
            in(x);in(y);
            query(1,1,n,x,y);
            cnt=0;
            For(j,1,t) if(vis[j]) cnt++;
            o(cnt);p('\n');
        }
    }
    return 0;
}

 

posted @ 2019-09-20 23:10  WeiAR  阅读(...)  评论(...编辑  收藏