SPOJ - UNTITLE1

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http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=22254

You are given a sequence of N integers A1, A2 .. AN. (-10000 <= Ai <= 10000, N <= 50000)

Let Si denote the sum of A1..Ai. You need to apply M (M <= 50000) operations:

0 x y k: increase all integers from Ax to Ay by k(1 <= x <= y <= N, -10000 <= k <= 10000).
1 x y: ask for max{ Si | x <= i <= y }.(1 <= x <= y <= N)

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对于区间增值[L,R,k]，

若 L<=i<=R，那么 S[i]=S[i]+k∗(i−L+1)=S[i]+k∗i+k∗(1−L)

若 i>R，那么 S[i]=S[i]+(R−L+1)∗k

可以写成这个形式： S[i]=X[i]∗i+Y[i]

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将序列分成N−−√ 块，每块记录两个标记 sumk 和 add，

表示对于这个块内元素都加值 sumk∗i+add

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对于区间加值的操作，修改整块的标记，对于不在整块的部分暴力修改即可。

对于询问操作，考虑得到单块的答案，观察 S[i]=sumk∗i+add，

可以发现，它可以写成斜率形式，可以用斜率优化来得到单块内的S[i]最大值，在凸包上二分斜率或者三分点即可。

但是斜率优化需要维护凸包，区间加值操作时对于修改了S[i]的块重新建造凸包即可。

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时间复杂度：O(M∗N−−√∗logN−−√)

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#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <string>#include <map>#include <vector>#include <stack>#include <queue>#include <utility>#include <iostream>#include <algorithm>template<class Num>void read(Num &x){    char c; int flag = 1;    while((c = getchar()) < '0' || c > '9')        if(c == '-') flag *= -1;    x = c - '0';    while((c = getchar()) >= '0' && c <= '9')        x = (x<<3) + (x<<1) + (c-'0');    x *= flag;    return;}template<class Num>void write(Num x){    if(x < 0) putchar('-'), x = -x;    static char s[20];int sl = 0;    while(x) s[sl++] = x%10 + '0',x /= 10;    if(!sl) {putchar('0');return;}    while(sl) putchar(s[--sl]);}const double eps = 1e-3;const int maxn = 50005, maxS = 300, maxB = 300;const long long LINF = 1LL<<60;#define REP(__x,__st,__ed) for(int __x = (__st); __x <= (__ed); __x++)int n, m, S, B, a[maxn];long long sum[maxn], sk[maxB], c[maxB];struct type_block{    int id, ll, rr;    int t[maxS], len;#define check(x,y,z) ((y - x)*(sum[z] - sum[y]) - (sum[y] - sum[x])*(z - y) >= 0)       void build()    {        len = 0;        for(int i = ll; i <= rr; i++)        {            while(len >= 2 && check(t[len-1], t[len], i)) t[len--] = 0;            t[++len] = i;        }    }#undef check    void init(int __id)    {        id = __id;        ll = (id - 1)*S + 1;        rr = std::min(id * S, n);        build();    }    long long calcu()    {        int l = 1, r = len;        long long c1, c2, ret = -LINF;        while(r - l >= 5)        {            int mid1 = (l + r)>>1, mid2 = (mid1 + r)>>1;            c1 = sum[t[mid1]] + t[mid1]*sk[id];            c2 = sum[t[mid2]] + t[mid2]*sk[id];            if(c1 < c2) l = mid1; else r = mid2;        }        for(int i = l; i <= r; i++)            ret = std::max(ret, sum[t[i]] + t[i]*sk[id]);        return ret + c[id];    }}block[maxB];int main(){#ifndef ONLINE_JUDGE        freopen("spoj.in","r",stdin);    freopen("spoj.out","w",stdout);#endif      read(n);    S = sqrt(n) + eps, B = n/S + ((n%S)?1:0);    REP(i, 1, n) read(a[i]), sum[i] = sum[i-1] + a[i];    REP(i, 1, B) block[i].init(i);    read(m);    REP(i, 1, m)    {        int op, l, r, k;        int l_blo, r_blo, st, ed;        read(op), read(l), read(r);        l_blo = (l - 1)/S + 1;        r_blo = (r - 1)/S + 1;        st = block[l_blo + 1].ll;        ed = block[r_blo - 1].rr;        if(!op)        {            read(k);            if(l_blo == r_blo)            {                REP(i, l, r) sum[i] += (i - l + 1)*k;                ed += S, ed = std::min(ed, n);                  REP(i, r + 1, ed) sum[i] += (r - l + 1)*k;                REP(i, r_blo + 1, B) c[i] += (r - l + 1)*k;                 block[l_blo].build();               }            else            {                REP(i, l, st - 1) sum[i] += (i - l + 1)*k;                REP(i, l_blo + 1, r_blo - 1)  sk[i] += k, c[i] += (1 - l)*k;                REP(i, ed + 1, r) sum[i] += (i - l + 1)*k;                ed += S, ed = std::min(ed, n);                  REP(i, r + 1, ed) sum[i] += (r - l + 1)*k;                REP(i, r_blo + 1, B) c[i] += (r - l + 1)*k;                 block[l_blo].build();                block[r_blo].build();            }        }        else        {            long long ans = -LINF;            if(l_blo == r_blo)            {                REP(i, l, r) ans = std::max(sum[i] + i*sk[l_blo] + c[l_blo], ans);            }            else            {                REP(i, l, st - 1) ans = std::max(sum[i] + i*sk[l_blo] + c[l_blo], ans);                REP(i, l_blo + 1, r_blo - 1) ans = std::max(block[i].calcu(), ans);                REP(i, ed + 1, r) ans = std::max(sum[i] + i*sk[r_blo] + c[r_blo], ans);            }            write(ans), puts("");        }    }#ifndef ONLINE_JUDGE        fclose(stdin);    fclose(stdout);#endif      return 0;}