zoukankan      html  css  js  c++  java
  • CF377D Developing Game

    题目链接:

    luogu

    题目分析:

    把每个人当成一个三元组([l_i, r_i, v_i])
    考虑每个人对哪个能力区间([L, R])有贡献
    应该是左端点在([l_i, v_i]),右端点在([v_i, r_i])的区间
    拍到一个二维平面上,求最多有多少个矩形交一起
    线段树维护扫描线即可

    代码:

    #include<bits/stdc++.h>
    #define N (600000 + 10)
    using namespace std;
    inline int read() {
    	int cnt = 0, f = 1; char c = getchar();
    	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
    	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
    	return cnt * f;
    }
    
    int n, Y[N], cnt, ans = 0, ansx, ansy, tot;
    int l[N], r[N], v[N];
    struct node2 {
    	int x, Y1, Y2, tag;
    }seg[N << 1];
    
    struct node {
    	int l, r;
    	int tag, gmax, pos;
    	#define l(p) tree[p].l
    	#define r(p) tree[p].r
    	#define tag(p) tree[p].tag
    	#define gmax(p) tree[p].gmax
    	#define pos(p) tree[p].pos
    } tree[N << 2];
    
    void pushdown(int p) {
    	tag(p << 1) += tag(p);
    	tag(p << 1 | 1) += tag(p);
    	gmax(p << 1) += tag(p);
    	gmax(p << 1 | 1) += tag(p);
    	tag(p) = 0;
    }
    
    void pushup(int p) {
    	if (gmax(p << 1) > gmax(p << 1 | 1)) {
    		pos(p) = pos(p << 1);
    		gmax(p) = gmax(p << 1);
    	} else {
    		pos(p) = pos(p << 1 | 1);
    		gmax(p) = gmax(p << 1 | 1);
    	}
    }
    
    void build (int p, int l, int r) {
    	l(p) = l, r(p) = r;
    	if (l == r) {pos(p) = Y[l]; return;}
    	int mid = (l + r) >> 1;
    	build (p << 1, l, mid);
    	build (p << 1 | 1, mid + 1, r);
    	pos(p) = pos(p << 1);
    }
    
    void modify(int p, int l, int r, int k) {
    	pushdown(p);
    	if (l <= Y[l(p)] && r >= Y[r(p)]) {tag(p) += k, gmax(p) += k; return;}
    	int mid = (l(p) + r(p)) >> 1;
    	if (l <= Y[mid]) modify(p << 1, l, r, k);
    	if (r > Y[mid]) modify(p << 1 | 1, l, r, k);
    	pushup(p);
    }
    bool cmp(node2 a, node2 b) {
    	return a.x == b.x ? a.tag > b.tag : a.x < b.x;
    }
    
    int main() {
    // 	freopen("data.in", "r", stdin);
    // 	freopen("myself.out", "w", stdout);
    	n = read();
    	for (register int i = 1; i <= n; ++i) {
    		l[i] = read(), v[i] = read(), r[i] = read();
    		seg[++tot].x = l[i], seg[tot].Y1 = v[i], seg[tot].Y2 = r[i], seg[tot].tag = 1, Y[tot] = v[i];
    		seg[++tot].x = v[i], seg[tot].Y1 = v[i], seg[tot].Y2 = r[i], seg[tot].tag = -1, Y[tot] = r[i];
    	}
    	sort (Y + 1, Y + tot + 1);
    	for (register int i = 1; i <= tot; ++i) if (Y[i] != Y[i + 1]) Y[++cnt] = Y[i];
    	build (1, 1, cnt);
    	sort (seg + 1, seg + tot + 1, cmp);
    //	for (register int i = 1; i <= cnt; ++i) cout<<Y[i]<<" "; return 0;
    	for (register int i = 1; i <= tot; ++i) {
    		modify(1, seg[i].Y1, seg[i].Y2, seg[i].tag);
    //		cout<<gmax(1)<<" ";
    		if (gmax(1) > ans) {
    			ans = gmax(1);
    			ansx = seg[i].x;
    			ansy = pos(1);
    		}
    	}
    //	return 0;
    	printf("%d
    ", ans);
    //	cout<<ansx<<" "<<ansy<<endl; return 0;
    	for (register int i = 1; i <= n; ++i) 
    		if (v[i] <= ansy && r[i] >= ansy && l[i] <= ansx && v[i] >= ansx) printf("%d ", i);
    	return 0;
    }
    
  • 相关阅读:
    一些C++11语言新特性
    项目管理计划应该包括哪些内容
    真相令人震惊!为什么越有钱的人,欠的钱越多?
    80后小伙返乡创业种植中药材,带领乡亲们脱贫致富
    Tableau
    知识点汇总
    决策树分析、EMV(期望货币值)
    信息系统项目管理师60天冲刺复习计划,2019下半年高项冲刺计划
    【系统分析师之路】系统分析师备考计划
    有一种规律:“劣币驱逐良币”,“坏人淘汰好人”(深度)
  • 原文地址:https://www.cnblogs.com/kma093/p/11791979.html
Copyright © 2011-2022 走看看