poj 2796

来源：互联网发布：oracle数据库基本语句编辑：程序博客网时间：2024/06/05 06:21

如果确定了 ax 为最小值，那么包含 ax 并且以 ax 为最小值的区间个数是 O(N2) 的。

因为序列中所有数的值非负，所以只需要考虑满足以上条件的最长区间即可。

而包含 ax 并且以 ax 为最小值的最长区间可以用两次单调队列求出，最后对每个 ax 计算答案即可。

#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]);}#define REP(__i,__st,__ed) for(int __i = (__st); __i <= (__ed); __i++)#define _REP(__i,__st,__ed) for(int __i = (__st); __i >= (__ed); __i--)const int maxn = 1e5 + 50;int n, a[maxn], pre[maxn], sur[maxn];int stack[maxn], top;long long ans, sum[maxn];int ansl, ansr;void init(){    read(n);    REP(i, 1, n) read(a[i]);}void prework(){    REP(i, 1, n) sum[i] = sum[i - 1] + a[i];    REP(i, 1, n)    {        while(top && a[i] <= a[stack[top]]) stack[top--] = 0;        pre[i] = (top?stack[top]:0) + 1, stack[++top] = i;    }    top = 0;    _REP(i, n, 1)    {        while(top && a[i] <= a[stack[top]]) stack[top--] = 0;        sur[i] = (top?stack[top]:(n + 1)) - 1, stack[++top] = i;    }}void solve(){    ans = a[1], ansl = ansr = 1;    REP(i, 1, n)    {        long long calc = (sum[sur[i]] - sum[pre[i] - 1])*a[i];        if(calc > ans) ans = calc, ansl = pre[i], ansr = sur[i];    }    write(ans), puts("");    write(ansl), putchar(' '), write(ansr);}int main(){#ifndef ONLINE_JUDGE    freopen("2796.in","r",stdin);    freopen("2796.out","w",stdout);#endif    init();    prework();    solve();#ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);#endif    return 0;           }

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