LA 3983 Robotruck

这道题感觉挺吃力的，还用到了我不熟悉的优先队列

题目中的推导也都看明白了，总之以后还要多体会才是

 

这里用优先对列的原因就是因为要维护一个滑动区间的最小值，比如在区间里２在１的前面，２在离开这个滑动区间会一直被１给“压迫”着，所以这个２是无用的，应当及时删除。这样的话剩下的元素都是递增的，优先队列也称单调队列也因此得名。

 

先把代码贴上：

 

 1 //#define LOCAL

 2 #include <iostream>

 3 #include <cstdio>

 4 #include <cstring>

 5 using namespace std;

 6 

 7 const int maxn = 100000 + 10;

 8 int x[maxn], y[maxn];

 9 int total_dist[maxn], total_weight[maxn], dist2origin[maxn];

10 int q[maxn], d[maxn];

11 

12 int func(int i)

13 {

14     return d[i] - total_dist[i+1] + dist2origin[i+1];

15 }

16 

17 int main(void)

18 {

19     #ifdef LOCAL

20         freopen("3983in.txt", "r", stdin);

21     #endif

22 

23     int T;

24     scanf("%d", &T);

25     while(T--)

26     {

27         int c, n, w, front, rear;

28         scanf("%d%d", &c, &n);

29         total_dist[0] = total_weight[0] = x[0] = y[0] = 0;

30         for(int i = 1; i <= n; ++i)

31         {

32             scanf("%d%d%d", &x[i], &y[i], &w);

33             dist2origin[i] = x[i] + y[i];

34             total_dist[i] = total_dist[i-1] + abs(x[i] - x[i-1]) + abs(y[i] - y[i-1]);

35             total_weight[i] = total_weight[i-1] + w;

36         }

37         front = rear = 1;

38         for(int i = 1; i <= n; ++i)

39         {

40             while(front <= rear && total_weight[i] - total_weight[q[front]] > c)

41                 ++front;

42             d[i] = func(q[front]) + total_dist[i] + dist2origin[i];

43             while(front <= rear && func(q[rear]) >= func(i))

44                 --rear;

45             q[++rear] = i;

46         }

47         printf("%d\n", d[n]);

48         if(T)    printf("\n");

49     }

50     return 0;

51 }

代码君