2716: [Violet 3]天使玩偶
Time Limit: 80 Sec Memory Limit: 128 MB
Submit: 1473 Solved: 621
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样例过大，略
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Source
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
时间： 2025-01-13 15:24:47