hdu 5023 A Corrupt Mayor's Performance Art（线段树+位运算）

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5023

解题思路：

题目大意：

一开始所有结点的颜色均为2类，然后给你两种操作：

P a b c 

将a到b区间全部图为c类

Q a b

输出a到b区间所有点的颜色种类。

算法思想：

线段树区间更新，用位运算存储区间颜色状态。。。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define INF 0xfffffffusing namespace std;const int N = 1000010;struct node{    int l,r;    int setv;    int color;}tree[N<<2];void build(int id,int l,int r){//id:index    tree[id].l = l;    tree[id].r = r;    tree[id].setv = 0;    tree[id].color = 2;    if(l == r)        return;    int mid = (l+r)>>1;    build(id<<1,l,mid);    build(id<<1|1,mid+1,r);}void pushup(int id){    tree[id].color = tree[id<<1].color | tree[id<<1|1].color;}void pushdown(int id){    if(tree[id].setv){        int tmp = tree[id].setv;        tree[id<<1].color = (1<<(tmp-1));        tree[id<<1|1].color = (1<<(tmp-1));        tree[id<<1].setv = tree[id<<1|1].setv = tmp;        tree[id].setv = 0;    }}void update(int id,int l,int r,int val){    if(tree[id].l >= l && tree[id].r <= r){        tree[id].setv = val;        tree[id].color = (1<<(val-1));        return;    }    pushdown(id);    int mid = (tree[id].l+tree[id].r)>>1;    if(l <= mid)        update(id<<1,l,r,val);    if(mid < r)        update((id<<1)|1,l,r,val);    pushup(id);}int query(int id,int l,int r){     if(tree[id].l >= l && tree[id].r <= r){        return tree[id].color;    }    pushdown(id);    int mid = (tree[id].l+tree[id].r)>>1;    int ans = 0;    if(l <= mid)        ans |= query(id<<1,l,r);    if(mid < r)        ans |= query(id<<1|1,l,r);    pushup(id);    return ans;}int main(){    int n,m;    while(scanf("%d%d",&n,&m),n+m){        build(1,1,n);        char op[10];        int a,b,c;        while(m--){            scanf("%s",op);            if(op[0] == 'P'){                scanf("%d%d%d",&a,&b,&c);                update(1,a,b,c);            }            else{                scanf("%d%d",&a,&b);                int ans = query(1,a,b);                int flag = 1;                for(int i = 0; i < 30; i++){                    if(ans & (1<<i)){                        if(flag == 1)                            printf("%d",i+1);                        else                            printf(" %d",i+1);                        flag++;                    }                }                printf("\n");            }        }    }    return 0;}