BZOJ 2716: [Violet 3]天使玩偶

2716: [Violet 3]天使玩偶

Time Limit: 80 Sec  Memory Limit: 128 MB
Submit: 1473  Solved: 621
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Description

Input

Output

Sample Input & Output

样例过大，略

HINT

 

 

Source

Vani原创 欢迎移步 OJ2648

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CDQ分治，分类讨论拆绝对值的方式，分别查询最优值。

 

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <algorithm>
  4 
  5 inline int nextChar(void) {
  6     const int siz = 1024;
  7     static char buf[siz];
  8     static char *hd = buf + siz;
  9     static char *tl = buf + siz;
 10     if (hd == tl)
 11         fread(hd = buf, 1, siz, stdin);
 12     return int(*hd++);
 13 }
 14 
 15 inline int nextInt(void) {
 16     register int ret = 0;
 17     register int neg = false;
 18     register int bit = nextChar();
 19     for (; bit < 48; bit = nextChar())
 20         if (bit == '-')neg ^= true;
 21     for (; bit > 47; bit = nextChar())
 22         ret = ret * 10 + bit - 48;
 23     return neg ? -ret : ret;
 24 }
 25 
 26 const int lim = 1000001;
 27 const int siz = 1000005;
 28 const int inf = 1000000007;
 29 
 30 int n, m;
 31 
 32 struct data {
 33     int k, x, y, t, p;
 34     inline friend bool operator < 
 35     (const data &a, const data &b) {
 36         if (a.x != b.x)
 37             return a.x < b.x;
 38         else
 39             return a.k < b.k;
 40     }
 41 }p[siz], s[siz], q[siz];
 42 
 43 int ans[siz];
 44 
 45 int now;
 46 int bit[siz];
 47 int tim[siz];
 48 
 49 inline void add(int t, int k) {
 50     for (; t < siz; t += t&-t)
 51         if (bit[t] < k || tim[t] != now)
 52             bit[t] = k, tim[t] = now;
 53 }
 54 
 55 inline int ask(int t) {
 56     int ret = -inf;
 57     for (; t; t -= t&-t)
 58         if (ret < bit[t] && tim[t] == now)
 59             ret = bit[t];
 60     return ret;
 61 }
 62 
 63 void cdqSolve(int l, int r) {
 64     if (l >= r)return;
 65     
 66     int mid = (l + r) >> 1;
 67     
 68     cdqSolve(l, mid);
 69     cdqSolve(mid + 1, r);
 70     
 71     int t1 = l, t2 = mid + 1, tot = l;
 72     
 73     while (t1 <= mid && t2 <= r) {
 74         if (s[t1] < s[t2])
 75             q[tot++] = s[t1++];
 76         else
 77             q[tot++] = s[t2++];
 78     }
 79     
 80     while (t1 <= mid)
 81         q[tot++] = s[t1++];
 82         
 83     while (t2 <= r)
 84         q[tot++] = s[t2++];
 85         
 86     for (int i = l; i <= r; ++i)
 87         s[i] = q[i];
 88     
 89     ++now;
 90     
 91     for (int i = l; i <= r; ++i)
 92         if (s[i].k && s[i].t > mid) {
 93             int tmp = s[i].x + s[i].y - ask(s[i].y);
 94             if (ans[s[i].p] > tmp)
 95                 ans[s[i].p] = tmp;
 96         }
 97         else if (!s[i].k && s[i].t <= mid)
 98             add(s[i].y, s[i].x + s[i].y);
 99 }
100 
101 inline void cdqSolve1(void) {
102     memcpy(s, p, sizeof(s));
103     cdqSolve(1, n + m);
104 }
105 
106 inline void cdqSolve2(void) {
107     memcpy(s, p, sizeof(s));
108     for (int i = 1; i <= n + m; ++i)
109         s[i].x = lim - s[i].x;
110     cdqSolve(1, n + m);
111 }
112 
113 inline void cdqSolve3(void) {
114     memcpy(s, p, sizeof(s));
115     for (int i = 1; i <= n + m; ++i)
116         s[i].y = lim - s[i].y;
117     cdqSolve(1, n + m);
118 }
119 
120 inline void cdqSolve4(void) {
121     memcpy(s, p, sizeof(s));
122     for (int i = 1; i <= n + m; ++i)
123         s[i].x = lim - s[i].x,
124         s[i].y = lim - s[i].y;
125     cdqSolve(1, n + m);
126 }
127 
128 signed main(void) {
129 //    freopen("in", "r", stdin);
130 //    freopen("out", "w", stdout);
131     
132     n = nextInt();
133     m = nextInt();
134     
135     for (int i = 1; i <= n; ++i) {
136         p[i].x = nextInt();
137         p[i].y = nextInt();
138         p[i].k = 0;
139         p[i].t = i;
140     }
141     
142     for (int i = 1; i <= m; ++i) {
143         p[i + n].k = nextInt() - 1;
144         p[i + n].x = nextInt();
145         p[i + n].y = nextInt();
146         p[i + n].t = i + n;
147         p[i + n].p = i;
148     }
149     
150     for (int i = 1; i <= m; ++i)
151         ans[i] = inf;
152     
153     cdqSolve1();
154     cdqSolve2();
155     cdqSolve3();
156     cdqSolve4();
157     
158     for (int i = 1; i <= m; ++i)
159         if (p[i + n].k)printf("%d\n", ans[i]);
160 }

 

@Author: YouSiki

posted @ 2017-01-03 16:10  YouSiki  阅读(734)  评论(0)  编辑  收藏  举报