HDU5023 - A Corrupt Mayor's Performance Art(线段树)

来源：互联网发布：阿里云数据中心编辑：程序博客网时间：2024/05/21 11:32

题目链接： HDU5023

【题意】给出一条线段有n个点，初始每个点的颜色都是2；给出两种操作：

‘Q’查询区间[a,b]内的所有包含的颜色从小到大输出；

‘P’给区间[a,b]染上c色；

【分析】线段树区间更新，区间查询，位运算记录下就好了

【AC CODE】218ms

#include <cstdio>#include <cstring>#include <vector>using namespace std;#define MAXN 1000010int color[MAXN<<2],vis[MAXN<<2],col;inline void update_up(int rt){color[rt] = color[rt<<1]|color[rt<<1|1];}void update_down(int rt){if(vis[rt]){vis[rt<<1] = vis[rt<<1|1] = vis[rt];color[rt<<1] = color[rt<<1|1] = vis[rt];vis[rt] = 0;}}void bulid(int l, int r, int rt){vis[rt] = 0;color[rt] = 2;if(l == r) return;int mid = (l+r)>>1;bulid(l,mid,rt<<1);bulid(mid+1,r,rt<<1|1);}void update(int l, int r, int c, int ql, int qr,int rt){if(ql <= l && qr >= r){color[rt] = vis[rt] = 1<<(c-1);return;}update_down(rt);int mid = (l+r)>>1;if(ql <= mid)update(l,mid,c,ql,qr,rt<<1);if(qr > mid)update(mid+1,r,c,ql,qr,rt<<1|1);update_up(rt);}int query(int l, int r, int ql, int qr, int rt){if(ql <= l && qr >= r)return color[rt];update_down(rt);int mid = (l+r)>>1, ans = 0;if(ql <= mid) ans |= query(l,mid,ql,qr,rt<<1);if(qr > mid) ans |= query(mid+1,r,ql,qr,rt<<1|1);return ans;}void output(){vector<int> ans;for(int i = 0; i < 30; i++){if(col&(1<<i))ans.push_back(i+1);}printf("%d", ans[0]);for(int i = 1; i < ans.size(); i++)printf(" %d", ans[i]);putchar('\n');}int main(){#ifdef SHYfreopen("e:\\1.txt","r",stdin);#endifint n,m;while(~scanf("%d %d%*c", &n, &m) && n+m){int a,b,c;char ch;bulid(1,n,1);for(int i = 0; i < m; i++){ch = getchar();if('P' == ch){scanf("%d %d %d%*c", &a, &b, &c);update(1,n,c,a,b,1);}else{scanf("%d %d%*c", &a, &b);col = query(1,n,a,b,1);output();}}}return 0;}

0 0