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

把求和操作改为或操作,就可以了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #define lson l,m,rt<<1
 7 #define rson m+1,r,rt<<1|1
 8 using namespace std;
 9 const int maxn=1111111;
10 int sum[maxn<<2];
11 int col[maxn<<2];
12 void up(int rt){
13     sum[rt]=sum[rt<<1]|sum[rt<<1|1];
14 }
15 void down(int rt){
16     if(col[rt]){
17         col[rt<<1]=col[rt<<1|1]=col[rt];
18         sum[rt<<1]=sum[rt<<1|1]=sum[rt];
19         col[rt]=0;
20     }
21 }
22 void build(int l,int r,int rt){
23     col[rt]=0;
24     sum[rt]=4;
25     if(l==r)return;
26     int m=(l+r)>>1;
27     build(lson);
28     build(rson);
29     up(rt);
30 }
31 void update(int L,int R,int c,int l,int r,int rt){
32     if(L<=l&&R>=r){
33         col[rt]=c;
34         sum[rt]=c;
35         return;
36     }
37     down(rt);
38     int m=(l+r)>>1;
39     if(L<=m)update(L,R,c,lson);
40     if(R>m)update(L,R,c,rson);
41     up(rt);
42 }
43 int query(int L,int R,int l,int r,int rt){
44     if(L<=l&&R>=r)return sum[rt];
45     down(rt);
46     int m=(l+r)>>1;
47     int ans=0;
48     if(L<=m)ans|=query(L,R,lson);
49     if(m<R)ans|=query(L,R,rson);
50     return ans;
51 }
52 int main()
53 {
54 //    freopen("in","r",stdin);
55     int n,q;
56     while(scanf("%d%d",&n,&q)>0&&(n|q)){
57         build(1,n,1);
58         char ch[5];
59         int a,b,c;
60         while(q--){
61             scanf("%s",ch);
62             if(ch[0]==‘P‘){
63                 scanf("%d%d%d",&a,&b,&c);
64                 update(a,b,1<<c,1,n,1);
65             }
66             else {
67                 scanf("%d%d",&a,&b);
68                 int ans=query(a,b,1,n,1);
69                 int fir=0;
70                 for(int i=1;i<31;i++){
71                     if(ans&(1<<i)){
72                         if(fir)printf(" ");
73                         printf("%d",i);
74                         fir=1;
75                     }
76                 }
77                 puts("");
78             }
79         }
80     }
81     return 0;
82 }

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

时间: 2024-10-25 05:15:39

hdu 5023 A Corrupt Mayor's Performance Art (线段树)的相关文章

HDU 5023 A Corrupt Mayor&#39;s Performance Art 线段树区间更新+状态压缩

Link:  http://acm.hdu.edu.cn/showproblem.php?pid=5023 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <vector> 6 #include <string> 7 #include <cmath> 8 using namesp

hdu 5023 A Corrupt Mayor&#39;s Performance Art(线段树区间更新)

#include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std; int tree[5001000],add[5001000]; int color[50]; int n,m; void pushup(int pos) { tree[pos]=tree[pos<<1]|tree[pos<<1|1]; //更新

hdu 5023 A Corrupt Mayor&#39;s Performance Art(线段树)

题目链接 题意:有一个长度 n 的序列,初始染色2,有两种操作,P x ,y ,z,区间x---y染色为z,另一种Q x,y,查询区间 x -- y 有几种颜色,并输出,会覆盖 分析:lz[]为0,表示下面颜色不统一,统一是>0; f[]表示该节点下有多少种颜色,是30种颜色的二进制表示. 刚开始做时,用1<<i 不对,不知道为什么,int的范围按理不会超的.. 1 #include <iostream> 2 #include <cstdio> 3 #includ

hdu 5023 A Corrupt Mayor&#39;s Performance Art (线段树+区间更新+状压)

A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others) Total Submission(s): 699    Accepted Submission(s): 267 Problem Description Corrupt governors always find ways to get dirty money.

HDU 5023 A Corrupt Mayor&#39;s Performance Art(线段树区间更新)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5023 解题报告:一面墙长度为n,有N个单元,每个单元编号从1到n,墙的初始的颜色是2,一共有30种颜色,有两种操作: P a b c  把区间a到b涂成c颜色 Q a b 查询区间a到b的颜色 线段树区间更新,每个节点保存的信息有,存储颜色的c,30种颜色可以压缩到一个int型里面存储,然后还有一个tot,表示这个区间一共有多少种颜色. 对于P操作,依次往下寻找,找要更新的区间,找到要更新的区间之前

HDU 5023 A Corrupt Mayor&#39;s Performance Art (线段树)

A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others) Total Submission(s): 255    Accepted Submission(s): 114 Problem Description Corrupt governors always find ways to get dirty money.

ACM学习历程—HDU 5023 A Corrupt Mayor&#39;s Performance Art(广州赛区网赛)(线段树)

Problem Description Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this seemed a safe way for mayor X to make money. Becaus

2014 网选 广州赛区 hdu 5023 A Corrupt Mayor&#39;s Performance Art

1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #define N 1000005 6 using namespace std; 7 8 int c[35]; 9 int tree[N*4];//正值表示该节点所管理的区间的颜色是纯色,-1表示的是非纯色 10 int n, m; 11 12 void buildT(int ld, int

HDU - 5023 A Corrupt Mayor&#39;s Performance Art(区间修改)

Problem Description Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this seemed a safe way for mayor X to make money. Becaus