• 洛谷P4344 [SHOI2015]脑洞治疗仪(ODT)


    题意

    题目链接

    Sol

    ODT板子题。

    操作1直接拆区间就行。

    #include<bits/stdc++.h>
    #define fi first
    #define se second 
    const int MAXN = 2e5 + 10;
    using namespace std;
    inline int read() {
    	char c = getchar(); int x = 0, f = 1;
    	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    	return x * f;
    }
    int N, M;
    #define sit set<Node>::iterator 
    struct Node {
    	int l, r;
    	mutable int v;
    	bool operator < (const Node &rhs) const {
    		return l < rhs.l;
    	}
    };
    set<Node> s;
    sit split(int p) {
    	sit pos = s.lower_bound({p, 0, 0});
    	if(pos->l == p) return pos;
    	pos--;
    	int L = pos->l, R = pos->r, V = pos->v;
    	s.erase(pos);
    	s.insert({L, p - 1, V});
    	return s.insert({p, R, V}).fi;
    }
    void Mem(int l, int r) {
    	sit ed = split(r + 1), bg = split(l);
    	s.erase(bg, ed);
    	s.insert({l, r, 0});
    }
    void Fix(int l0, int r0, int l, int r) {
    	int num = 0;
    	sit ed = split(r + 1), bg = split(l);
    	for(sit i = bg; i != ed; i++) if(i->v == 1) num += i->r - i->l + 1;
    	s.erase(bg, ed);
    	s.insert({l, r, 0});
    	ed = split(r0 + 1), bg = split(l0);
    	sit gg; int RR = -1;
    	for(sit i = bg; i != ed; i++) {
    		if(i -> v == 1) continue;
    		if(num <= 0) return ;
    		int len = i->r - i->l + 1;
    		if(len <= num) {i -> v = 1, num -= len; gg = i; RR = i->r; continue;}
    		int L = i->l, R = i->r;
    		s.erase(i);
    		s.insert({L, L + num - 1, 1});
    		s.insert({L + num, R, 0});
    		return ;
    	}
    	if(RR > l0) {//Ò»¸ö²¢Ã»ÓÐʲôÂÑÓõÄÓÅ»¯ 
    		gg++;
    		s.erase(bg, gg);
    		s.insert({l0, RR, 1});	
    	}
    }
    int Query(int l, int r) {
    	sit ed = split(r + 1), bg = split(l);
    	int pre = 0, ans = 0;
    	for(sit i = bg; i != ed; i++) {
    		if(i->v == 0) ans = max(ans, i->r - i->l + 1 + pre), pre += i->r - i->l + 1;
    		else pre = 0;
    	}
    	return ans;
    }
    int main() {
    	//freopen("a.in", "r", stdin);
    	N = read(); M = read();
    	for(int i = 1; i <= N; i++) s.insert({i, i, 1});
    	s.insert({N + 1, N + 1, 1});
    	for(int i = 1; i <= M; i++) {
    		int opt = read(), l = read(), r = read();
    		if(opt == 0) Mem(l, r);
    		else if(opt == 1) {
    			int l1 = read(), r1 = read();
    			Fix(l1, r1, l, r);
    		} else printf("%d
    ", Query(l, r));
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10357218.html
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