BZOJ3476 [Usaco2014 Mar]The Lazy Cow

这题做的我真是23333。。。

奥，这题表述不清，是说曼哈顿距离 <= k

首先我们坐标变换：

设(X, Y)为原坐标，则新坐标为(X + Y, X - Y + E)，为了防止y < 0

然后我们只要维护一堆边平行于坐标轴的正方形即可

方法是扫描线扫过去，然后用线段树维护y轴上的权值，也就是维护段+1和求全部的最大值操作。

然后发现Rank2。。。为了Rank1开始丧心病狂，写起了入读优化、、、

发现差4ms就Rank1，极为不爽。。。

于是czr大爷告诉我可以fread。。。一番折腾后。。。

Rank1！欧液！

 

 1 /**************************************************************
 2     Problem: 3476
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:2192 ms
 7     Memory:64480 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cctype>
12 #include <algorithm>
13  
14 #define lson p << 1
15 #define rson p << 1 | 1
16 #define Tag seg[p].tag
17 using namespace std;
18 const int N = 100005;
19 const int E = 1000001;
20 const int Height = (E << 1) - 1;
21 const int M = (E << 3) + 10;
22  
23 struct point {
24     int x, y, v;
25 }a[N];
26 inline bool operator < (const point &a, const point &b) {
27     return a.x < b.x;
28 }
29  
30 struct segment_tree {
31     int tag, v;
32 }seg[M];
33  
34 int n, k;
35 int L, R, del;
36  
37 inline unsigned int read() {
38     int x = 0;
39     char ch = getchar();
40     while (ch < '0' || '9' < ch)
41         ch = getchar();
42     while ('0' <= ch && ch <= '9') {
43         x = x * 10 + ch - '0';
44         ch = getchar();
45     }
46     return x;
47 }
48  
49 inline void refresh(int p, int v) {
50     seg[p].v += v;
51     seg[p].tag += v;
52 }
53  
54 void insert(int p, int l, int r) {
55     if (L <= l && r <= R) {
56         refresh(p, del);
57         return;
58     }
59     if (Tag) {
60         refresh(lson ,Tag), refresh(rson, Tag);
61         Tag = 0;
62     }
63     int mid = l + r >> 1;
64     if (L <= mid)
65         insert(lson, l, mid);
66     if (mid < R) 
67         insert(rson, mid + 1, r);
68     seg[p].v = max(seg[lson].v, seg[rson].v);
69 }
70  
71 int main() {    
72     int i, j, X, Y, ans = 0;
73     n = read(), k = read();
74     k = k * 2 + 1;
75     for (i = 1; i <= n; ++i) {
76         a[i].v = read(), X = read(), Y = read();
77         a[i].x = X + Y, a[i].y = X - Y + E;
78     }
79     sort(a + 1, a + n + 1);
80     for (i = j = 1; i <= n; ++i) {
81         while (a[i].x - a[j].x + 1 > k) {
82             L = a[j].y - k + 1, R = a[j].y, del = -a[j].v;
83             insert(1, 1, Height), ++j;
84         }
85         L = a[i].y - k + 1, R = a[i].y, del = a[i].v;
86         insert(1, 1, Height);
87         ans = max(ans, seg[1].v);
88     }
89     printf("%d\n", ans);
90     return 0;
91 }
View Code(正常读入优化版)

↓下面这就是那个坑爹fread的版本。。。：

 

  1 /**************************************************************
  2     Problem: 3476
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:1940 ms
  7     Memory:66336 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cctype>
 12 #include <algorithm>
 13  
 14 #define lson p << 1
 15 #define rson p << 1 | 1
 16 #define Tag seg[p].tag
 17 using namespace std;
 18 const int N = 100005;
 19 const int E = 1000001;
 20 const int Height = (E << 1) - 1;
 21 const int M = (E << 3) + 10;
 22 const int MaxCH = 1900100;
 23 struct point {
 24     int x, y, v;
 25 }a[N];
 26 inline bool operator < (const point &a, const point &b) {
 27     return a.x < b.x;
 28 }
 29  
 30 struct segment_tree {
 31     int tag, v;
 32 }seg[M];
 33  
 34 int n, k;
 35 int L, R, del;
 36 char CH[MaxCH];
 37 int len, Left = 0;
 38  
 39 inline unsigned int read() {
 40     int x = 0;
 41     char ch = getchar();
 42     while (ch < '0' || '9' < ch)
 43         ch = getchar();
 44     while ('0' <= ch && ch <= '9') {
 45         x = x * 10 + ch - '0';
 46         ch = getchar();
 47     }
 48     return x;
 49 }
 50  
 51 inline int find() {
 52     int x = 0;
 53     while (CH[Left] < '0' || '9' < CH[Left])
 54         ++ Left;
 55     while ('0' <= CH[Left] && CH[Left] <= '9')
 56         x = x * 10 + CH[Left++] - '0';
 57     return x;
 58 }
 59  
 60 inline void refresh(int p, int v) {
 61     seg[p].v += v;
 62     seg[p].tag += v;
 63 }
 64  
 65 void insert(int p, int l, int r) {
 66     if (L <= l && r <= R) {
 67         refresh(p, del);
 68         return;
 69     }
 70     if (Tag) {
 71         refresh(lson, Tag), refresh(rson, Tag);
 72         Tag = 0;
 73     }
 74     int mid = l + r >> 1;
 75     if (L <= mid)
 76         insert(lson, l, mid);
 77     if (mid < R) 
 78         insert(rson, mid + 1, r);
 79     seg[p].v = max(seg[lson].v, seg[rson].v);
 80 }
 81  
 82 int main() {    
 83     int i, j, X, Y, ans = 0;
 84     len = fread(CH, 1, MaxCH, stdin);
 85     CH[len] = ' ';
 86     n = find(), k = find();
 87     k = k * 2 + 1;
 88     for (i = 1; i <= n; ++i) {
 89         a[i].v = find(), X = find(), Y = find();
 90         a[i].x = X + Y, a[i].y = X - Y + E;
 91     }
 92     sort(a + 1, a + n + 1);
 93     for (i = j = 1; i <= n; ++i) {
 94         while (a[i].x - a[j].x + 1 > k) {
 95             L = a[j].y - k + 1, R = a[j].y, del = -a[j].v;
 96             insert(1, 1, Height), ++j;
 97         }
 98         L = a[i].y - k + 1, R = a[i].y, del = a[i].v;
 99         insert(1, 1, Height);
100         ans = max(ans, seg[1].v);
101     }
102     printf("%d\n", ans);
103     return 0;
104 }
View Code(fread坑爹版)

 

posted on 2014-11-16 15:39  Xs酱~  阅读(495)  评论(0)  编辑  收藏  举报