【BZOJ1645】[Usaco2007 Open]City Horizon 城市地平线【线段树】【矩形面积并】

【题目链接】

/* Pigonometry */
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 40005, maxm = maxn << 1;

typedef long long LL;

int n, m, pos[maxm], tr[maxm << 2], mxv[maxm << 2];

struct _line {
	int l, r, w;
} line[maxn];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline int find(int x) {
	int l = 1, r = m;
	while(l <= r) {
		int mid = l + r >> 1;
		if(pos[mid] < x) l = mid + 1;
		else if(pos[mid] == x) return mid;
		else r = mid - 1;
	}
}

inline void pushup(int p) {
	tr[p] = max(tr[p << 1], tr[p << 1 | 1]);
}

inline void pushdown(int p) {
	if(mxv[p]) {
		int l = p << 1, r = p << 1 | 1;
		mxv[l] = max(mxv[l], mxv[p]);
		mxv[r] = max(mxv[r], mxv[p]);
		tr[l] = max(tr[l], mxv[p]);
		tr[r] = max(tr[r], mxv[p]);
	}
}

inline void change(int p, int l, int r, int x, int y, int c) {
	if(x <= l && r <= y) {
		tr[p] = max(tr[p], c);
		mxv[p] = max(mxv[p], c);
		return;
	}
	int mid = l + r >> 1;
	pushdown(p);
	if(x <= mid) change(p << 1, l, mid, x, y, c);
	if(y > mid) change(p << 1 | 1, mid + 1, r, x, y, c);
	pushup(p);
}

inline int query(int p, int l, int r, int x) {
	if(l == r && r == x) return tr[p];
	int mid = l + r >> 1;
	pushdown(p);
	if(x <= mid) return query(p << 1, l, mid, x);
	else return query(p << 1 | 1, mid + 1, r, x);
}

int main() {
	n = iread(); m = n << 1;

	for(int i = 1; i <= n; i++) {
		int l = iread(), r = iread(), w = iread();
		line[i] = (_line){l, r, w};
		pos[(i << 1) - 1] = l; pos[i << 1] = r;
	}

	sort(pos + 1, pos + 1 + m);
	m = unique(pos + 1, pos + 1 + m) - (pos + 1);
	for(int i = 1; i <= n; i++) {
		line[i].l = find(line[i].l);
		line[i].r = find(line[i].r);
		change(1, 1, m, line[i].l, line[i].r - 1, line[i].w);
	}

	LL ans = 0;
	for(int i = 1; i < m; i++) 
		ans += (LL)(pos[i + 1] - pos[i]) * query(1, 1, m, i);

	printf("%lld\n", ans);
	return 0;
}


你可能感兴趣的:(矩形面积并)