HDU 5023 A Corrupt Mayor's Performance Art (据说是线段树)

Posted 2020-08-19 dwtfukgv

题意：给定一个1-n的墙，然后有两种操作，一种是P l ,r, a 把l-r的墙都染成a这种颜色，另一种是 Q l, r 表示，输出 l-r 区间内的颜色。

析：应该是一个线段树+状态压缩，但是我用set暴力过去了。用线段树+状态压缩，区间更新，很简单，就不说了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int mod = 1e9 + 7;
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
    int l, r;
    Node() { }
    Node(int ll, int rr) : l(ll), r(rr) { }
    bool operator < (const Node &p) const{
        return r < p.r;
    }
};
set<Node> sets[35];
set<Node> :: iterator it, it1;

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        if(!m && !n)  break;
        for(int i = 1; i <= 30; ++i)  sets[i].clear();
        sets[2].insert(Node(1, n));
        char s[5];
        int a, b, c;
        Node u;
        while(m--){
            scanf("%s", s);
            if(s[0] == ‘P‘){
                scanf("%d %d %d", &a, &b, &c);
                for(int i = 1; i <= 30; ++i){
                    if(sets[i].size() == 0)  continue;
                    while(true){
                        it1 = sets[i].lower_bound(Node(0, a));
                        if(it1 == sets[i].end() || it1->l > b)  break;
                        u = *it1;
                        sets[i].erase(it1);
                        if(u.l < a){
                            sets[i].insert(Node(u.l, a-1));
                            if(u.r > b) sets[i].insert(Node(b+1, u.r));
                        }
                        else if(u.r > b) sets[i].insert(Node(b+1, u.r));
                    }
                }
                sets[c].insert(Node(a, b));
            }
            else{
                scanf("%d %d", &a, &b);
                int cnt = 0;
                for(int i = 1; i < 31; ++i){
                    if(sets[i].size() == 0)  continue;
                    it1 = sets[i].lower_bound(Node(0, a));
                    if(it1 == sets[i].end() || it1->l > b) continue;
                    if(cnt)  putchar(‘ ‘);
                    printf("%d", i);
                    ++cnt;
                }
                putchar(‘\n‘);
            }
        }
    }
    return 0;
}

 线段树：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL sum[maxn<<2], sets[maxn<<2];

void pushup(int rt){
    sum[rt] = sum[rt<<1] | sum[rt<<1|1];
}

void pushdown(int rt, int len){
    int l = rt<<1, r = rt<<1|1;
    if(sets[rt]){
        sum[r] = sum[l] = sets[rt];
        sets[l] = sets[r] = sets[rt];
        sets[rt] = 0;
    }
}

void build(int l, int r, int rt){
    sets[rt] = 0;
    if(l == r){
        sum[rt] = 2;
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int L, int R, int val, int l, int r, int rt){
    if(L <= l && R >= r){
        sets[rt] = 1<<val-1;
        sum[rt] = 1<<val-1;
        return ;
    }
    pushdown(rt, r-l+1);
    int m = (l + r) >> 1;
    if(L <= m)  update(L, R, val, lson);
    if(R > m)   update(L, R, val, rson);
    pushup(rt);
}

LL query(int L, int R, int l, int r, int rt){
    if(L <= l && R >= r) return sum[rt];
    pushdown(rt, r-l+1);
    LL ans = 0;
    int m = (l + r) >> 1;
    if(L <= m)  ans |= query(L, R, lson);
    if(m < R)   ans |= query(L, R, rson);
    return ans;
}

int main(){
    while(scanf("%d %d", &n, &m) == 2 && m+n){
        build(1, n, 1);
        int l, r, val;
        char s[5];
        while(m--){
            scanf("%s", s);
            if(s[0] == ‘P‘){
                scanf("%d %d %d", &l, &r, &val);
                update(l, r, val, 1, n, 1);
            }
            else{
                scanf("%d %d", &l, &r);
                LL ans = query(l, r, 1, n, 1);
                int cnt = 0;
                for(int i = 0; i < 30; ++i) if(ans & (1<<i)){
                    if(cnt)  putchar(‘ ‘);
                    printf("%d", i+1);
                    ++cnt;
                }
                printf("\n");
            }
        }
    }
    return 0;
}