CSU 1529 - Equator(双端队列)

题目：

Description

Input

Output

Sample Input

33 1 2 38 4 5 -1 -1 1 -1 -1 52 -1 -1

Sample Output

6140

题意：

循环序列，求出最大的子串和。

思路：

双端队列，。

维护一个不大于n的队列，队尾减队首更新答案。

AC.

#include <iostream>#include <cstdio>#include <cmath>using namespace std;const int INF = 1e5;const int maxn = 1e6+5;int a[maxn*2], sum[maxn*2];int deq[maxn*2];int main(){    //freopen("in", "r", stdin);    int T;    scanf("%d", &T);    while(T--) {        int n;        scanf("%d", &n);        sum[0] = 0;        for(int i = 1; i <= n; ++i) {            scanf("%d", &a[i]);            a[i+n] = a[i];            sum[i] = sum[i-1] + a[i];        }        for(int i = n+1; i <= 2*n; ++i) {            sum[i] = sum[i-1] + a[i];        }        int ans = 0;        deq[0] = 0;        int s = 0, t = -1;        for(int i = 1; i <= 2*n; ++i) {            while(s <= t && sum[i] < sum[deq[t]]) {                ans = max(ans, sum[deq[t]] - sum[deq[s]]);                t--;            }            while(s <= t && i - deq[s] > n) {                ans = max(ans, sum[deq[t]] - sum[deq[s]]);                s++;            }            deq[++t] = i;        }        ans = max(ans, sum[deq[t]] - sum[deq[s]]);        printf("%d\n", ans);    }    return 0;}