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

时间：2023-03-08 18:07:56

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;

}

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