【线段树】 HDU 5025 A Corrupt Mayor's Performance Art

更新区间内颜色

输出区间内的颜色总数

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <math.h>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define IN     freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 5040;//点数的最大值
const int MAXM = 20006;//边数的最大值
const int INF = 0x3f3f3f3f;
const int mod = 10000007;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid (l+r)>>1
#define maxn 1100000
int sum[maxn<<2],col[maxn<<2];
int num[34];
void pushup(int rt) //向上更新
{
    sum[rt]=sum[rt<<1]|sum[rt<<1|1];
}
void pushdown(int rt)
{
    if(col[rt]) //向下更新
    {
        col[rt<<1]=col[rt<<1|1]=col[rt];
        sum[rt<<1]=sum[rt<<1|1]=sum[rt];
        col[rt]=0;
    }
}
void build(int l,int r,int rt)
{
    sum[rt]=num[2];//初始为颜色1
    if(l==r)
    {
        return;
    }
    int m=mid;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        sum[rt]=num[c];
        col[rt]=num[c];
        return;
    }
    pushdown(rt);
    int m=mid;
    if(L<=m) update(L,R,c,lson);
    if(R>m) update(L,R,c,rson);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        return sum[rt];
    }
    pushdown(rt);
    int m=mid;
    int res=0;
    if(L<=m) res=res|query(L,R,lson);
    if(R>m) res=res|query(L,R,rson);
    return res;
}
int main()
{
    int n,q;
    memset(num,0,sizeof(num));
    num[1]=1;
    for(int i=2; i<=30; i++) //先储存每种颜色。。。
        num[i]=num[i-1]<<1;
    //IN;
    while(scanf("%d%d ",&n,&q),n+q)
    {
        build(1,n,1);
        while(q--)
        {
            char s[3];
            scanf("%s",s);
            int a,b,c;
            if(s[0]=='Q')
            {
                int ans[33],wei=1,point=0;
                scanf("%d%d",&a,&b);
                int x=query(a,b,1,n,1);
                while(x)//统计颜色个数
                {
                    if(x&1==1)
                        ans[point++]=wei;
                    x=x>>1;
                    wei++;
                }
                for(int i=0;i<point;i++)
                    printf("%d%c",ans[i],i==point-1?'\n':' ');
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                update(a,b,c,1,n,1);
            }
        }
    }
    return 0;
}

【线段树】 HDU 5025 A Corrupt Mayor's Performance Art

时间: 2024-10-05 06:41:25

【线段树】 HDU 5025 A Corrupt Mayor's Performance Art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)

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

HDU 5023 A Corrupt Mayor&#39;s Performance Art(线段树+优美的位运算)

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

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)

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 ma

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)

题意:一长由N小段组成的长条,每小段的初始颜色为2.现执行M个操作,每个操作是以下两种中的一种(0 < N <= 1,000,000; 0<M<=100,000) : P a b c --> 将段a到段b涂成颜色c,c是1, 2, ... 30中的一种(0 < a<=b <= N, 0 < c <= 30). Q a b --> 问段a到段b之间有哪几种颜色,按颜色大小从小到大输出(0 < a<=b <= N, 0 <