URAL 2014

题目大意是一个人出国留学只用信用卡消费，会出现某天有金钱变化，这些都只会有两种情况，一种是挣钱，一种是花钱，如果钱不够就需要用信用卡透支。这些天还会像家里的父亲寄一封信来说明当天支出和收入。但是由于快递不行，所以到达父亲手里的信封的时间是乱序的，问每当收到一封信，父亲能推算的最大透支额度是多少。

思路：

线段树

先把时间离散化，然后对每一封信，用线段树更新当前天到最晚天的收支，最后用线段树维护当前所有天的最小值即是结果。

#include 
using namespace std;
int n;
const int maxn = 100000 + 10;
struct Letter
{
	int val, t;
}L[maxn];
map id;
const int month[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int daysec = 24 * 60 * 60;
typedef long long ll;
ll sumv[maxn<<2], minv[maxn<<2], pos[maxn], add[maxn<<2];
void pushdown(int o, int l, int r) {
	if(add[o]) {
		int m = (l+r)>>1;
		add[o<<1] += add[o];
		add[o<<1|1] += add[o];
		sumv[o<<1] += add[o] * (m - l + 1);
		sumv[o<<1|1] += add[o] * (r - m);
		minv[o<<1] += add[o];
		minv[o<<1|1] += add[o];
		add[o] = 0;
	}
}
void pushup(int o, int l, int r) {
	sumv[o] = sumv[o<<1] + sumv[o<<1|1];
	minv[o] = min(minv[o<<1], minv[o<<1|1]);
}
ll qsum(int o, int l, int r, int ql, int qr) {
	if(ql <= l && qr >= r) {
		return sumv[o];
	}
	pushdown(o, l, r);
	int m = (l+r)>>1;
	ll sum = 0;
	if(ql <= m) sum += qsum(o<<1, l, m, ql, qr);
	if(qr > m)  sum += qsum(o<<1|1, m + 1, r, ql, qr);
	return sum;
}
ll qmin(int o, int l, int r, int ql, int qr) {
	if(ql <= l && qr >= r) {
		return min(0ll, minv[o]);
	}
	pushdown(o, l, r);
	int m = (l+r)>>1;
	ll minv = 0;
	if(ql <= m) minv = min(minv, qmin(o<<1, l, m, ql, qr));
	if(qr > m) minv = min(minv, qmin(o<<1|1, m + 1, r, ql, qr));
	return minv;
}
void update(int o, int l, int r, int ql, int qr, int V) {
	if(ql <= l && qr >= r) {
		sumv[o] += (ll)V * (r - l + 1);
		minv[o] += (ll)V;
		add[o] += (ll)V;
		return;
	}
	pushdown(o, l, r);
	int m = (l+r)>>1;
	if(ql <= m) update(o<<1, l, m, ql, qr, V);
	if(qr > m) update(o<<1|1, m + 1, r, ql, qr, V);
	pushup(o, l, r);
}
int main(int argc, char const *argv[])
{
	scanf("%d", &n);
	vector T;
	for(int i = 0; i < n; ++i) {
		int d, m, h, s;
		scanf("%d %d.%d %d:%d", &L[i].val, &d, &m, &h, &s);
		int sec= s * 60 + h * 3600;
		int day = 0;
		for(int j = 1; j < m; ++j) day += month[j];
		day += d;
		sec += daysec * day;
		L[i].t = sec;
		T.push_back(sec);
	}
	sort(T.begin(), T.end());
	int cnt = 0;
	for(int i = 0; i < (int)T.size(); ++i) {
		id[T[i]] = ++cnt;
	}
	for(int i = 0; i < n; ++i) {
		L[i].t = id[L[i].t];
		pos[L[i].t] = i;
	}
	for(int i = 0; i < n; ++i) {
		int val = L[i].val, t = L[i].t;
			update(1, 1, n, t, n, val);
		printf("%lld\n", qmin(1, 1, n, 1, n));
	}
	return 0;
}