HDU 3016 Man Down(线段树)

Posted 2020-09-20 claireyuancy

HDU 3016 Man Down

题目链接

题意：是男人就下100层的游戏的简单版，每次仅仅能从两端下落。求落地最大血量

思路：利用线段树能够处理出每一个线段能来自哪几个线段。然后就是dag最长路了

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 100005;

int n;

struct Line {
	int l, r, y, val;
	Line() {}
	Line(int l, int r, int y, int val) {
		this->l = l; this->r = r;
		this->y = y; this->val = val;
	}
	void read() {
		scanf("%d%d%d%d", &y, &l, &r, &val);
	}
} line[N];

bool cmp(Line a, Line b) {
	return a.y < b.y;
}

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)

struct Node {
	int l, r, id, lazy;
	void gao(int v) {
		lazy = v;
		id = v;
	}
} node[4 * N];

void build(int l, int r, int x = 0) {
	node[x].l = l; node[x].r = r;
	node[x].id = node[x].lazy = -1;
	if (l == r) return;
	int mid = (l + r) / 2;
	build(l, mid, lson(x));
	build(mid + 1, r, rson(x));
}

void pushdown(int x) {
	if (node[x].lazy != -1) {
		node[lson(x)].gao(node[x].lazy);
		node[rson(x)].gao(node[x].lazy);
		node[x].lazy = -1;
	}
}

void add(int l, int r, int v, int x = 0) {
	if (node[x].l >= l && node[x].r <= r) {
		node[x].gao(v);
		return;
	}
	pushdown(x);
	int mid = (node[x].l + node[x].r) / 2;
	if (l <= mid) add(l, r, v, lson(x));
	if (r > mid) add(l, r, v, rson(x));
}

int query(int v, int x = 0) {
	if (node[x].l == node[x].r) return node[x].id;
	int mid = (node[x].l + node[x].r) / 2;
	pushdown(x);
	if (v <= mid) return query(v, lson(x));
	if (v > mid) return query(v, rson(x));
}

const int INF = 0x3f3f3f3f;

int dp[N];

vector<int> g[N];

int main() {
	while (~scanf("%d", &n)) {
		build(0, 100000);
		line[n] = Line(0, 100000, 0, 0);
		for (int i = 0; i < n; i++) {
			line[i].read();
			g[i].clear();
		}
		n++;
		sort(line, line + n, cmp);
		for (int i = 0; i < n; i++) {
			if (i) {
				int tol = query(line[i].l);
				int tor = query(line[i].r);
				g[tol].push_back(i);
				if (tol != tor)
					g[tor].push_back(i);
			}
			add(line[i].l, line[i].r, i);
		}
		line[n - 1].val += 100;
		dp[n - 1] = line[n - 1].val;
		for (int i = n - 2; i >= 0; i--) {
			dp[i] = -INF;
			for (int j = 0; j < g[i].size(); j++)
				dp[i] = max(dp[i], dp[g[i][j]] + line[i].val);
		}
		if (dp[0] <= 0) dp[0] = -1;
		printf("%d\n", dp[0]);
	}
	return 0;
}