码农题.
LEN 开到 1000 比较稳.
1 #include <cstdio>
2 #include <cstring>
3 #include <cstdlib>
4 #include <cctype>
5 #include <algorithm>
6 #include <vector>
7 using namespace std;
8 #define F(i, a, b) for (register int i = (a); i <= (b); i++)
9 #define LL long long
10 inline int rd(void) {
11 int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -1;
12 int x = 0; for (; isdigit(c); c = getchar()) x = x*10+c-‘0‘; return x*f;
13 }
14
15 const int N = 50005;
16 const int Ni = 2005;
17 const int Mi = 4005;
18 const int S = 100005;
19 const int INF = ~0u>>2;
20 const int LEN = 1000;
21
22 int n, Start[N];
23
24 int List[Ni], tmp[Ni]; bool Vis[Ni];
25
26 struct Edge { int v, f, nx; }mp[Mi]; int tot, hd[Ni];
27 int q[Ni], qh, qt, Lev[Ni];
28 inline void Init(int x, int y, int w) {
29 mp[++tot] = (Edge){y, w, hd[x]}, hd[x] = tot;
30 mp[++tot] = (Edge){x, w, hd[y]}, hd[y] = tot;
31 }
32
33 struct E { int v, d; }; vector<E> g[S];
34 inline void Add(int x, int y, int w) { g[x].push_back((E){y, w}), g[y].push_back((E){x, w}); }
35
36 bool BFS(int s, int t) {
37 memset(q, 0, sizeof q), qh = qt = 0, memset(Lev, -1, sizeof Lev);
38 q[++qt] = s, Lev[s] = 0;
39 while (qh != qt) {
40 int x = q[++qh];
41 for (int k = hd[x]; k > 0; k = mp[k].nx)
42 if (mp[k].f > 0 && Lev[mp[k].v] == -1) {
43 Lev[mp[k].v] = Lev[x]+1;
44 if (mp[k].v == t) return true;
45 q[++qt] = mp[k].v;
46 }
47 }
48 return false;
49 }
50 int DFS(int x, int t, int flow) {
51 if (x == t) return flow;
52 int sum = 0;
53 for (int k = hd[x]; k > 0; k = mp[k].nx)
54 if (mp[k].f > 0 && Lev[mp[k].v] == Lev[x]+1) {
55 int tmp = DFS(mp[k].v, t, min(flow, mp[k].f));
56 if (tmp > 0) {
57 mp[k].f -= tmp, mp[k^1].f += tmp;
58 sum += tmp, flow -= tmp;
59 if (!flow) break;
60 }
61 else Lev[mp[k].v] = INF;
62 }
63 return sum;
64 }
65 void Paint(int x) {
66 Vis[x] = true;
67 for (int k = hd[x]; k > 0; k = mp[k].nx)
68 if (mp[k].f > 0 && !Vis[mp[k].v])
69 Paint(mp[k].v);
70 }
71 void Solve(int Bg, int L, int R) {
72 if (L == R) return;
73
74 int s = List[L], t = List[R];
75 for (int i = 2; i <= tot; i += 2)
76 mp[i].f = mp[i^1].f = (mp[i].f + mp[i^1].f) >> 1;
77
78 int sum = 0;
79 while (BFS(s, t))
80 sum += DFS(s, t, INF);
81 memset(Vis, false, sizeof Vis);
82 Paint(s);
83
84 Add(Bg + s, Bg + t, sum);
85 F(i, L, R) tmp[i] = List[i];
86 int curL = L, curR = R;
87 F(i, L, R)
88 Vis[tmp[i]] ? List[curL++] = tmp[i] : List[curR--] = tmp[i];
89 Solve(Bg, L, curL-1);
90 Solve(Bg, curR+1, R);
91 }
92
93 void Build(int Bg, int n, int m) {
94 tot = 1, memset(hd, 0, sizeof hd);
95 F(i, 1, m) {
96 int x = rd(), y = rd(), w = rd();
97 Init(x, y, w);
98 }
99
100 F(i, 1, n) List[i] = i;
101 random_shuffle(List+1, List+n+1);
102 Solve(Bg, 1, n);
103 }
104
105 int val[N], id[N], ord[N];
106 struct point { int x, y; }p[N];
107
108 inline bool cmpx(int i, int j) { return p[i].x < p[j].x; }
109 inline bool cmpy(int i, int j) { return p[i].y < p[j].y; }
110 inline LL Dist(point A, point B) { return 1LL * (A.x - B.x) * (A.x - B.x) + 1LL * (A.y - B.y) * (A.y - B.y); }
111 void Up(void) {
112 F(i, 1, n) {
113 int L = max(1, i-LEN), R = min(n, i+LEN), cur = ord[i];
114 F(j, L, R) if (i != j) {
115 int Dx = Dist(p[cur], p[id[cur]]), Dy = Dist(p[cur], p[ord[j]]);
116 if (Dx > Dy || (Dx == Dy && id[cur] > ord[j]))
117 id[cur] = ord[j];
118 }
119 }
120 }
121
122 int par[S], key[S], dep[S];
123 void Prework(int x) {
124 for (vector<E>::iterator it = g[x].begin(); it != g[x].end(); it++)
125 if (par[x] != it->v) {
126 if (par[it->v] > 0) { key[it->v] += it->d; continue; }
127 par[it->v] = x, key[it->v] = it->d, dep[it->v] = dep[x]+1;
128 Prework(it->v);
129 }
130 }
131 inline int Query(int x, int y) {
132 int res = INF;
133 if (dep[x] < dep[y]) swap(x, y);
134 for (; dep[x] != dep[y]; x = par[x])
135 res = min(res, key[x]);
136 for (; x != y; x = par[x], y = par[y])
137 res = min(res, key[x]), res = min(res, key[y]);
138 // If x and y is not connected, the answer should be zero
139 // and I can get zero through ( par[root] = 0, key[root] = 0 )
140 return res;
141 }
142
143 int main(void) {
144 #ifndef ONLINE_JUDGE
145 freopen("network.in", "r", stdin);
146 #endif
147
148 n = rd();
149 F(j, 1, n) {
150 int x = rd(); p[j] = (point){x, rd()};
151 val[j] = rd();
152
153 int ni = rd(), mi = rd();
154 Start[j+1] = Start[j] + ni;
155
156 Build(Start[j], ni, mi);
157 }
158
159 if (n != 1) {
160 F(i, 1, n) id[i] = 1 + (i == 1);
161 F(i, 1, n) ord[i] = i;
162 sort(ord+1, ord+n+1, cmpx);
163 Up();
164 sort(ord+1, ord+n+1, cmpy);
165 Up();
166 F(i, 1, n)
167 Add(Start[i]+1, Start[id[i]]+1, val[i]);
168 }
169
170 F(i, 1, Start[n+1])
171 if (!key[i]) dep[i] = 1, Prework(i);
172
173 int q = rd();
174 F(t, 1, q) {
175 int ta = rd(), tb = rd(), qa = rd(), qb = rd();
176 printf("%d\n", Query(Start[ta] + qa, Start[tb] + qb));
177 }
178
179 return 0;
180 }
时间： 2024-10-19 14:47:58