Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.
Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.
Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.
You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.
The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.
The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
3
10 10 5
5 7 2
5 12 4
5
30 25 20 15 10
9 10 12 4 13
9 20 35 11 25
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
题意 : 输入的一个 n ,表示天数,然后输入的第一行表示在这天产生的雪的量,然后再一行表示雪在这天的融化量是多少,对于当前的一天,他不仅会融化当天的,还会融化前面所有天的雪,最后输出每天融化雪的量。
思路分析 :最开始的想法是纯用线段树去搞它,写了一会交上去WA了... 后来问了下大佬的思路,才知道原来这题是去二分雪的量,去判断雪在哪天融化 。然后差分一遍就没问题了。
代码示例 :
#define ll long long const ll maxn = 1e5+5; const double pi = acos(-1.0); const ll inf = 0x3f3f3f3f; ll v[maxn], te[maxn]; ll sum[maxn]; ll ans[maxn], pp[maxn], f[maxn]; ll start; bool check(ll p, ll vi){ ll ss = sum[p] - sum[start-1]; if (ss >= vi) return true; else return false; } void fun(ll l, ll r, ll vi){ ll mid, pos; while(l <= r){ mid = (l+r) >> 1; if (check(mid, vi)) {pos = mid; r = mid-1;} else l = mid+1; } //printf("mid = %lld ", pos); ans[pos]--; ans[start]++; pp[pos] += vi-(sum[pos-1]-sum[start-1]); } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); ll n; cin >> n; for(ll i = 1; i <= n; i++) scanf("%d", &v[i]); for(ll i = 1; i <= n; i++) { scanf("%lld", &te[i]); sum[i] = sum[i-1] + te[i]; } sum[n+1] = sum[n] + 10000000000; for(ll i = 1; i <= n; i++){ start = i; fun(i, n+1, v[i]); } for(ll i = 1; i <= n; i++) { ans[i] += ans[i-1]; f[i] = ans[i]*te[i] + pp[i]; } for(ll i = 1; i <= n; i++) printf("%lld%c", f[i], i==n?' ':' '); return 0; }
方法二 : 优先队列 + 前缀和
看了大佬们的代码,发现优先队列也可以,每次将当天前面所有的温度和累加再加上当前天的雪的总量,这样打入优先队列的雪量就可以视为是从同一天打入的
代码示例 :
#define ll long long const ll maxn = 1e5+5; const double pi = acos(-1.0); const ll inf = 0x3f3f3f3f; ll v[maxn], te[maxn]; ll sum[maxn]; priority_queue<ll, vector<ll>, greater<ll> >que; int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); ll n; cin >> n; for(ll i = 1; i <= n; i++) scanf("%d", &v[i]); for(ll i = 1; i <= n; i++) { scanf("%lld", &te[i]); sum[i] = sum[i-1] + te[i]; } ll pt = 0; for(ll i = 1; i <= n; i++){ que.push(sum[i-1]+v[i]); pt += te[i]; ll ans = 0; while(!que.empty()){ ll x = que.top(); if (x > pt) break; que.pop(); ans += x-(pt-te[i]); } ans += que.size()*te[i]; printf("%lld%c", ans, i==n?' ':' '); } return 0; }