CDOJ1324-卿学姐与公主 【线段树点更新】

http://acm.uestc.edu.cn/#/problem/show/1324

卿学姐与公主

Time Limit: 2000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)
 
 

某日，百无聊赖的卿学姐打开了某11区的某魔幻游戏

在这个魔幻的游戏里，生活着一个美丽的公主，但现在公主被关押在了魔王的城堡中。

英勇的卿学姐拔出利刃冲向了拯救公主的道路。

走过了荒野，翻越了高山，跨过了大洋，卿学姐来到了魔王的第一道城关。

在这个城关面前的是魔王的精锐部队，这些士兵成一字排开。

卿学姐的武器每次只能攻击一个士兵，并造成一定伤害，卿学姐想知道某时刻从L

到R

这个区间内，从开始到现在累计受伤最严重的士兵受到的伤害。

最开始每个士兵的受到的伤害都是0

Input

第一行两个整数N,Q表示总共有N个士兵编号从1到N，和Q个操作。

接下来Q行，每行三个整数，首先输入一个t，如果t是1，那么输入p,x，表示卿学姐攻击了p这个位置的士兵，并造成了x的伤害。如果t是2，那么输入L,R，表示卿学姐想知道现在[L,R]闭区间内，受伤最严重的士兵受到的伤害。

1≤N≤100000

1≤Q≤100000

1≤p≤N

1≤x≤100000

1≤L≤R≤N1≤L≤R≤

Output

对于每个询问，回答相应的值

Sample input and output

Sample InputSample Output
 
5 4
2 1 2
1 2 4
1 3 5
2 3 3

0
5

 

思路：线段树 点更新

代码：

  1 #include <fstream>
  2 #include <iostream>
  3 using namespace std;
  4 
  5 #define MAX 100005
  6 #define ll long long int
  7 
  8 class SegmentTree
  9 {
 10 private:
 11     int n;
 12     struct Node
 13     {
 14         int left, right, mid;
 15         ll maxval;
 16         Node (): maxval(0)
 17         {}
 18     }*tree;
 19 public:
 20     SegmentTree (int n): n(n) 
 21     {
 22         tree = new Node[n<<2];
 23     }
 24     ~SegmentTree ()
 25     {
 26         delete []tree;
 27     }
 28     void Build ();
 29     void Update (int id, int subid, int val);
 30     ll Search (int id, int left, int right);
 31 };
 32 void SegmentTree::Build () //非递归，由于叶子只能出现在树的最后两层，故而可以以tree[i].left < n作为循环条件进行非递归操作。需要说明的是，这里把最后一层的多余内存也初始化了，不过不影响全局。
 33 {
 34     int t1;
 35     tree[1].left = 1;
 36     tree[1].right = n;
 37     tree[1].mid = (n + 1) >> 1;
 38     for (int i = 1; tree[i].left < n; i++)
 39     {
 40         if (tree[i].left < tree[i].right)
 41         {
 42             t1 = i << 1;
 43             tree[t1].left = tree[i].left;
 44             tree[t1].right = tree[i].mid;
 45             tree[t1].mid = (tree[t1].left + tree[t1].right) >> 1;
 46             t1++;
 47             tree[t1].left = tree[i].mid + 1;
 48             tree[t1].right = tree[i].right;
 49             tree[t1].mid = (tree[t1].left + tree[t1].right) >> 1;
 50         }
 51     }
 52 }
 53 void SegmentTree::Update (int id, int subid, int val)
 54 {
 55     if (tree[id].left == tree[id].right)
 56     {
 57         tree[id].maxval += val;
 58     }
 59     else if (tree[id].mid >= subid)
 60     {
 61         Update(id << 1, subid, val);
 62         tree[id].maxval = max(tree[id << 1].maxval, tree[id].maxval);
 63     }
 64     else if (tree[id].mid < subid)
 65     {
 66         Update((id << 1)|1, subid, val);
 67         tree[id].maxval = max(tree[id].maxval, tree[(id << 1)|1].maxval);
 68     }
 69 }
 70 ll SegmentTree::Search (int id, int left, int right)
 71 {
 72     if (tree[id].left == left && tree[id].right == right)
 73     {
 74         return tree[id].maxval;
 75     }
 76     else if (tree[id].mid >= right)
 77     {
 78         return Search(id << 1, left, right);
 79     }
 80     else if (tree[id].mid < left)
 81     {
 82         return Search((id << 1)|1, left, right);
 83     }
 84     else
 85     {
 86         return max(Search(id << 1, left, tree[id].mid), Search((id << 1)|1, tree[id].mid + 1, right));
 87     }
 88 }
 89 
 90 int main ()
 91 {
 92     //freopen("D:\\input.in","r",stdin);
 93     int n, q, t, p, x;
 94     scanf ("%d %d", &n, &q);
 95     SegmentTree lv(n);
 96     lv.Build();
 97     for (int i = 0; i < q; i++)
 98     {
 99         scanf ("%d %d %d", &t, &p, &x);
100         if (t == 1)
101         {
102             lv.Update(1, p, x);
103         }
104         else
105         {
106             printf("%lld\n", lv.Search(1, p, x));
107         }
108     }
109     return 0;
110 }

 代码2：递归

  1 #include <fstream>
  2 #include <iostream>
  3 using namespace std;
  4 
  5 #define MAX 100005
  6 #define ll long long int
  7 
  8 class SegmentTree
  9 {
 10 private:
 11     int n;
 12     struct Node
 13     {
 14         int left, right, mid;
 15         ll maxval;
 16         Node (): maxval(0)
 17         {}
 18     }*tree;
 19 public:
 20     SegmentTree (int n): n(n) 
 21     {
 22         tree = new Node[n<<2];
 23     }
 24     ~SegmentTree ()
 25     {
 26         delete []tree;
 27     }
 28     void Build (int id, int left, int right);
 29     void Update (int id, int subid, int val);
 30     ll Search (int id, int left, int right);
 31 };
 32 void SegmentTree::Build (int id, int left, int right) //递归
 33 {
 34     tree[id].left = left;
 35     tree[id].right = right;
 36     tree[id].mid = (left + right) >> 1;
 37     if (tree[id].mid == right) 
 38     {
 39         return;
 40     }
 41     Build(id << 1, left, tree[id].mid);
 42     Build((id << 1)|1, tree[id].mid + 1, right);
 43 }
 44 void SegmentTree::Update (int id, int subid, int val)
 45 {
 46     if (tree[id].left == tree[id].right)
 47     {
 48         tree[id].maxval += val;
 49     }
 50     else if (tree[id].mid >= subid)
 51     {
 52         Update(id << 1, subid, val);
 53         tree[id].maxval = max(tree[id << 1].maxval, tree[id].maxval);
 54     }
 55     else if (tree[id].mid < subid)
 56     {
 57         Update((id << 1)|1, subid, val);
 58         tree[id].maxval = max(tree[id].maxval, tree[(id << 1)|1].maxval);
 59     }
 60 }
 61 ll SegmentTree::Search (int id, int left, int right)
 62 {
 63     if (tree[id].left == left && tree[id].right == right)
 64     {
 65         return tree[id].maxval;
 66     }
 67     else if (tree[id].mid >= right)
 68     {
 69         return Search(id << 1, left, right);
 70     }
 71     else if (tree[id].mid < left)
 72     {
 73         return Search((id << 1)|1, left, right);
 74     }
 75     else
 76     {
 77         return max(Search(id << 1, left, tree[id].mid), Search((id << 1)|1, tree[id].mid + 1, right));
 78     }
 79 }
 80 
 81 int main ()
 82 {
 83     //freopen("D:\\input.in","r",stdin);
 84     int n, q, t, p, x;
 85     scanf ("%d %d", &n, &q);
 86     SegmentTree lv(n);
 87     lv.Build(1, 1, n);
 88     for (int i = 0; i < q; i++)
 89     {
 90         scanf ("%d %d %d", &t, &p, &x);
 91         if (t == 1)
 92         {
 93             lv.Update(1, p, x);
 94         }
 95         else
 96         {
 97             printf("%lld\n", lv.Search(1, p, x));
 98         }
 99     }
100     return 0;
101 }

 

posted @ 2016-05-11 09:38  jiu~  阅读(290)  评论(0)  编辑  收藏  举报