ZOJ 3606Lazy Salesgirl

Description

Kochiya Sanae is a lazy girl who makes and sells bread. She is an expert at bread making and selling. She can sell the i-th customer some pieces of bread at price p_i for each piece. But she is so lazy that she will fall asleep if no customer comes to buy bread for more than wminutes. When she is sleeping, the customer coming to buy bread will wake her up and leave without buying anything. Once she is woken up, she will start to sell bread again until she encounters another gap of w minutes. What's more weird, she can sell 1 + ((k - 1) mod 3) pieces of bread when she sells at the k-th time. It's known that she starts to sell bread now and the i-th customer comes after t_iminutes. What is the minimum possible value of w that maximizes the average value of the bread sold each time?

Input

There are multiple test cases. The first line of input is an integer T ≈ 100 indicating the number of test cases.

The first line of each test case contains an integer 1 ≤ n ≤ 10⁵ indicating the number of customers. The second line contains n integers 1 ≤p_i ≤ 10⁶. The third line contains n integers 1 ≤ t_i ≤ 10⁷. All t_i are different.

Output

For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.

Sample Input

241 2 3 41 3 6 1041 2 3 44 7 9 10

Sample Output

3.000000 4.6666673.000000 6.666667

线段树的一种巧妙的应用，同时让我想通了一个之前一直在思考的问题。

看到的题解一般都是第一种办法的

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int maxn = 1e5 + 10;int T, n, p[maxn];struct point{int p, x, t, id;}t[maxn];struct STree{LL f[maxn << 2][3], c[maxn << 2];void clear(){memset(f, 0, sizeof(f));memset(c, 0, sizeof(c));}void add(int x, int l, int r, int u){c[x]++;if (l == r){for (int i = 0; i < 3; i++) f[x][i] = (i + 1) * p[u];}else{int m = l + r >> 1;if (u <= m) add(x << 1, l, m, u);else add(x << 1 | 1, m + 1, r, u);for (int i = 0; i < 3; i++){f[x][i] = f[x << 1][i] + f[x << 1 | 1][(c[x << 1] + i) % 3];}}}}solve;bool cmp1(const point&a, const point&b){return a.x < b.x;}bool cmp2(const point&a, const point&b){return a.t < b.t;}int main(){scanf("%d", &T);while (T--){scanf("%d", &n);solve.clear();for (int i = 1; i <= n; i++) scanf("%d", &t[i].p);for (int i = 1; i <= n; i++) scanf("%d", &t[i].x);sort(t + 1, t + n + 1, cmp1);  t[0].x = 0;for (int i = 1; i <= n; i++) t[i].t = t[i].x - t[i - 1].x, t[i].id = i, p[i] = t[i].p;//solve.build(1, 1, n);sort(t + 1, t + n + 1, cmp2);double w = t[1].t, ans = 0;for (int i = 1, j = 1; i <= n; i = j){while (j <= n && t[j].t == t[i].t) solve.add(1, 1, n, t[j++].id);double now = 1.0*solve.f[1][0] / (j - 1);if (now > ans) ans = now, w = t[i].t;}printf("%.6lf %.6lf\n", w, ans);}return 0;}

第二种从后往前一个一个删除

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int maxn = 1e5 + 10;int T, n, p[maxn];struct point{int p, x, t, id;}t[maxn];struct STree{LL f[maxn << 4][3], c[maxn << 4];void update(int x){for (int i = 0; i < 3; i++){f[x][i] = f[x << 1][i] + f[x << 1 | 1][i];}}void build(int x, int l, int r){c[x] = 0;if (l == r){for (int i = 3; i; i--){f[x][3 - i] = t[l].p*((l + i - 1) % 3 + 1);}}else{int m = l + r >> 1;build(x << 1, l, m);build(x << 1 | 1, m + 1, r);update(x);}}void make(int x){for (int i = 0; i < 2; i++){swap(f[x][i], f[x][i + 1]);}}void pushdown(int x){c[x] %= 3;c[x << 1] += c[x];c[x << 1 | 1] += c[x];while (c[x]--){make(x << 1);make(x << 1 | 1);}c[x] = 0;}void del(int x, int l, int r, int u){if (l == r){for (int i = 0; i < 3; i++) f[x][i] = 0;}else{if (c[x]) pushdown(x);int m = l + r >> 1;if (u <= m){del(x << 1, l, m, u);c[x << 1 | 1]++;make(x << 1 | 1);}else del(x << 1 | 1, m + 1, r, u);update(x);}}}solve;bool cmp1(const point&a, const point&b){return a.x < b.x;}bool cmp2(const point&a, const point&b){return a.t > b.t;}int main(){scanf("%d", &T);while (T--){scanf("%d", &n);for (int i = 1; i <= n; i++) scanf("%d", &t[i].p);for (int i = 1; i <= n; i++) scanf("%d", &t[i].x);sort(t + 1, t + n + 1, cmp1);  t[0].x = 0;for (int i = 1; i <= n; i++) t[i].t = t[i].x - t[i - 1].x, t[i].id = i;solve.build(1, 1, n);sort(t + 1, t + n + 1, cmp2);double w, ans = 0;for (int i = 1, j = 1; i <= n; i = j){double now = 1.0*solve.f[1][0] / (n - i + 1);if (now > ans) ans = now, w = t[i].t;else if (now==ans) w=t[i].t;while (j <= n && t[j].t == t[i].t) solve.del(1, 1, n, t[j++].id);}printf("%.6lf %.6lf\n", w, ans);}return 0;}

温故而知新，这次从前面来
#include<map>#include<cmath>    #include<queue> #include<string>#include<vector>#include<cstdio>    #include<cstring>    #include<algorithm>    using namespace std;#define ms(x,y) memset(x,y,sizeof(x))    #define rep(i,j,k) for(int i=j;i<=k;i++)    #define per(i,j,k) for(int i=j;i>=k;i--)    #define loop(i,j,k) for (int i=j;i!=-1;i=k[i])    #define inone(x) scanf("%d",&x)    #define intwo(x,y) scanf("%d%d",&x,&y)    #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)    typedef long long LL;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 1e5 + 10;int T, n, p[N], t[N], a[N], b[N], c[N];int g[N << 2];double f[N << 2][3];void push(int x, int l, int r){g[l] += g[x]; g[r] += g[x];for (g[x] %= 3; g[x]; g[x]--){rep(i, 0, 1) swap(f[l][i], f[l][i + 1]);rep(i, 0, 1) swap(f[r][i], f[r][i + 1]);}}void insert(int x, int l, int r, int u, int v){if (l == r) { rep(i, 0, 2) f[x][i] = v + v * i;for (g[x] %= 3; g[x]; g[x]--)rep(i, 0, 1) swap(f[x][i], f[x][i + 1]);return; }if (g[x]) push(x, x << 1, x << 1 | 1);int mid = l + r >> 1;if (u <= mid) insert(x << 1, l, mid, u, v);else insert(x << 1 | 1, mid + 1, r, u, v);rep(i, 0, 2) f[x][i] = f[x << 1][i] + f[x << 1 | 1][i];}void rotate(int x, int l, int r, int ll, int rr){if (ll <= l && r <= rr){rep(i, 0, 1) swap(f[x][i], f[x][i + 1]);g[x]++; return;}if (g[x]) push(x, x << 1, x << 1 | 1);int mid = l + r >> 1;if (ll <= mid) rotate(x << 1, l, mid, ll, rr);if (rr > mid) rotate(x << 1 | 1, mid + 1, r, ll, rr);rep(i, 0, 2) f[x][i] = f[x << 1][i] + f[x << 1 | 1][i];}bool cmp(int x, int y) { return t[x] < t[y]; }int main(){for (inone(T); T--;){inone(n); t[0] = b[0] = 0;rep(i, 1, n) inone(p[i]);rep(i, 1, n) inone(t[i]), a[i] = b[i] = i;sort(b + 1, b + n + 1, cmp);per(i, n, 1) t[b[i]] = t[b[i]] - t[b[i - 1]];rep(i, 1, n) c[b[i]] = i;sort(a + 1, a + n + 1, cmp);ms(f, 0); ms(g, 0);double W = 0, A = 0;for (int i = 1, j; i <= n; i = j){for (j = i; j <= n&&t[a[i]] == t[a[j]]; j++){insert(1, 1, n, c[a[j]], p[a[j]]);if (c[a[j]] < n) rotate(1, 1, n, c[a[j]] + 1, n);}if (A*(j - 1) < f[1][0]) A = f[1][0] / (j - 1), W = t[a[i]];}printf("%.6lf %.6lf\n", W, A); }return 0;}