HDU 3415 Max Sum of Max-K-sub-sequence(单调队列)

/*AC
思路：使用sum[i]存储前i个序列之和，队列中存储区间内出现过最小序列和，这样只需要sum[i]前去最小序列和即可。
*/
#include <iostream>
using namespace std;
const int nMax = 100010;
const int INF = 0x7fffffff;
int A[nMax];
int sum[2 * nMax];//这里需要增倍
struct Queue
{
int pos;
int v;
Queue(){}
Queue(int pos, int v): pos(pos), v(v){}
}queue[2 * nMax];
int front, rear;
int t, n, k;
int main()
{
//freopen("e://data.txt", "r", stdin);
scanf("%d", &t);
while(t --)
{
scanf("%d%d", &n, &k);
sum[0] = 0;
for(int i = 1; i <= n; ++ i)
{
scanf("%d", &A[i]);
sum[i] = sum[i - 1] + A[i];
}
for(int i = n + 1; i <= n + k; ++ i)
sum[i] = sum[i - 1] + A[i - n];
front = rear = 0;
int Max = -INF;
int l, r;
queue[front ++] = Queue(0, 0);//队列中存储最小元素和
for(int i = 1; i <= n + k; ++ i)
{
while(rear < front && (i - queue[rear].pos) > k)//这里做了改动
rear ++;
if(sum[i] - queue[rear].v > Max)//把结果运算放到了前面，这样更好一些
{
Max = sum[i] - queue[rear].v;
l = queue[rear].pos + 1;
r = i;
}
while(rear < front && queue[front - 1].v > sum[i])
front --;
queue[front ++] = Queue(i, sum[i]);

}
printf("%d %d %d\n", Max, (l - 1) % n + 1, (r - 1) % n + 1);
}
return 0;
}

/*AC
错误出在①处
#include <iostream>
using namespace std;
const int nMax = 100010;
int A[nMax];
int sum[2 * nMax];//这里需要增倍
struct Queue
{
int pos;
int v;
Queue(){}
Queue(int pos, int v): pos(pos), v(v){}
}queue[2 * nMax];
int front, rear;
int t, n, k;
int main()
{
//freopen("e://data.txt", "r", stdin);
scanf("%d", &t);
while(t --)
{
scanf("%d%d", &n, &k);
sum[0] = 0;
for(int i = 1; i <= n; ++ i)
{
scanf("%d", &A[i]);
sum[i] = sum[i - 1] + A[i];
}
for(int i = n + 1; i <= n + k; ++ i)
sum[i] = sum[i - 1] + A[i - n];
front = rear = 0;
int Max = sum[1];//初始化
int l, r;
l = r = 1;
queue[front ++] = Queue(0, 0);//队列中存储最小元素和
for(int i = 1; i < n + k; ++ i)
{
while(rear < front && queue[front - 1].v > sum[i])
front --;
queue[front ++] = Queue(i, sum[i]);
while(rear < front && (i + 1 - queue[rear].pos) > k)//①这个需要在if之前进行判断
rear ++;
if(sum[i + 1] - queue[rear].v > Max)//[queue[rear].pos, i + 1]的最大值
{
Max = sum[i + 1] - queue[rear].v;
l = queue[rear].pos + 1;
r = i + 1;
}
}
printf("%d %d %d\n", Max, (l - 1) % n + 1, (r - 1) % n + 1);
}
return 0;
}
*/

/*
第一次求解思路：
队列中存储长度在k之内的最大和，并存储l,r，在执行入队时候，需要遍历一次队列寻找出相连的元素，出队时判断l是否越界。
超时！

#include <iostream>
using namespace std;
const int INF = 0x7fffffff;
const int nMax = 100010;
int A[nMax];
struct Queue
{
int l, r;
int sum;
Queue(){}
Queue(int l, int r, int sum):l(l), r(r), sum(sum){}
}queue[nMax], ans;
int front, rear;
int t, n, k;
int main()
{
//freopen("e://data.txt", "r", stdin);
scanf("%d", &t);
while(t --)
{
scanf("%d%d", &n, &k);
int i;
for(i = 0; i < n; ++ i)
scanf("%d", &A[i]);
front = rear = 0;
int max;
int l;
for(i = 0; i < k; ++ i)
{
max = A[i];
l = i;
for(int j = rear; j < front; ++ j)
if(queue[j].r == i - 1 && queue[j].sum + A[i] > max)
{
max = queue[j].sum + A[i];
l = queue[j].l;
}
while(rear < front && queue[front - 1].sum < max)
front --;
queue[front ++] = Queue(l, i, max);
ans = queue[rear];
}
for(int j = i; j < n + k; ++ j)
{
while(rear < front && queue[rear].l < j - k + 1) rear ++;
i = j % n;
max = A[i];
l = i;
for(int j = rear; j < front; ++ j)
if(((i == 0 && queue[j].r == n - 1) || queue[j].r == i - 1) && queue[j].sum + A[i] > max)
{
max = queue[j].sum + A[i];
l = queue[j].l;
}
while(rear < front && queue[front - 1].sum < max)
front --;
queue[front ++] = Queue(l, i, max);
if(ans.sum < max)
ans = queue[rear];
}
printf("%d %d %d\n", ans.sum, ans.l + 1, ans.r + 1);
}
return 0;
}
*/