洛谷P3233 世界树
作者：互联网
题意：给定树上k个关键点，每个点属于离他最近，然后编号最小的关键点。求每个关键点管辖多少点。
解：虚树 + DP。
虚树不解释。主要是DP。用二元组存虚树上每个点的归属和距离。这一部分是二次扫描与换根法。
然后把关键点改为虚树节点，统计每个虚树节点管辖多少个节点，用SIZ表示，初始时SIZ = siz，SIZ[RT] = n。
如果一条虚树边两端点的归属相同。那么SIZ[fa] -= siz[son]
否则树上倍增找到y是最靠上属于的son的，然后SIZ[fa] -= siz[y] SIZ[son] = siz[y]
1 #include <cstdio>
2 #include <algorithm>
3 #include <vector>
4 #include <cstring>
5
6 typedef long long LL;
7 const int N = 300010, INF = 0x3f3f3f3f;
8
9 struct Edge {
10 int nex, v, len;
11 }edge[N * 2], EDGE[N * 2]; int tp, TP;
12
13 struct Node {
14 int x, d;
15 Node(int X = 0, int D = 0) {
16 x = X;
17 d = D;
18 }
19 }small[N];
20
21 int e[N], E[N], RT, siz[N], d[N], num, pos[N], pw[N], Time, imp[N], stk[N], top, now[N], imp2[N];
22 int ans[N], fr[N], SIZ[N], n, fa[N][20], use[N];
23
24 inline void add(int x, int y) {
25 tp++;
26 edge[tp].v = y;
27 edge[tp].nex = e[x];
28 e[x] = tp;
29 return;
30 }
31
32 inline void ADD(int x, int y) {
33 // printf("ADD %d %d \n", x, y);
34 TP++;
35 EDGE[TP].v = y;
36 EDGE[TP].len = d[y] - d[x];
37 EDGE[TP].nex = E[x];
38 E[x] = TP;
39 return;
40 }
41
42 inline bool cmp(const int &a, const int &b) {
43 return pos[a] < pos[b];
44 }
45
46 void DFS_1(int x, int father) {
47 fa[x][0] = father;
48 d[x] = d[father] + 1;
49 siz[x] = 1;
50 pos[x] = ++num;
51 for(int i = e[x]; i; i = edge[i].nex) {
52 int y = edge[i].v;
53 if(y == father) {
54 continue;
55 }
56 DFS_1(y, x);
57 siz[x] += siz[y];
58 }
59 return;
60 }
61
62 inline int lca(int x, int y) {
63 if(d[x] > d[y]) {
64 std::swap(x, y);
65 }
66 int t = pw[n];
67 while(t >= 0 && d[x] < d[y]) {
68 if(d[fa[y][t]] >= d[x]) {
69 y = fa[y][t];
70 }
71 t--;
72 }
73 if(x == y) {
74 return x;
75 }
76 t = pw[n];
77 while(t >= 0 && fa[x][0] != fa[y][0]) {
78 if(fa[x][t] != fa[y][t]) {
79 x = fa[x][t];
80 y = fa[y][t];
81 }
82 t--;
83 }
84 return fa[x][0];
85 }
86
87 inline void clear(int x) {
88 if(use[x] != Time) {
89 use[x] = Time;
90 E[x] = 0;
91 }
92 return;
93 }
94
95 inline void build_t(int k) {
96 std::sort(imp + 1, imp + k + 1, cmp);
97 TP = top = 0;
98 clear(imp[1]);
99 stk[++top] = imp[1];
100 for(int i = 2; i <= k; i++) {
101 int x = imp[i], y = lca(stk[top], x);
102 clear(x); clear(y);
103 while(top > 1 && pos[y] <= pos[stk[top - 1]]) {
104 ADD(stk[top - 1], stk[top]);
105 top--;
106 }
107 if(y != stk[top]) {
108 ADD(y, stk[top]);
109 stk[top] = y;
110 }
111 stk[++top] = x;
112 }
113 while(top > 1) {
114 ADD(stk[top - 1], stk[top]);
115 top--;
116 }
117 RT = stk[top];
118 return;
119 }
120
121 void out_t(int x) {
122 printf("out x = %d \n", x);
123 for(int i = E[x]; i; i = EDGE[i].nex) {
124 int y = EDGE[i].v;
125 // printf("EDGE %d y %d \n", i, y);
126 out_t(y);
127 }
128 return;
129 }
130
131 void getSmall(int x) {
132 (now[x] == Time) ? small[x] = Node(x, 0) : small[x] = Node(n + 1, INF);
133 // printf("getSmall x = %d small = %d \n", x, small[x].x);
134 SIZ[x] = siz[x];
135 for(int i = E[x]; i; i = EDGE[i].nex) {
136 int y = EDGE[i].v;
137 getSmall(y);
138 if(small[x].d > small[y].d + EDGE[i].len) {
139 small[x] = small[y];
140 small[x].d += EDGE[i].len;
141 }
142 else if(small[x].d == small[y].d + EDGE[i].len) {
143 small[x].x = std::min(small[x].x, small[y].x);
144 }
145 }
146 return;
147 }
148
149 void getEXsmall(int x, Node t) {
150 // printf("EX x = %d small = %d \n", x, small[x].x);
151 if(small[x].d > t.d || (small[x].d == t.d && small[x].x > t.x)) {
152 small[x] = t;
153 }
154 // printf("x = %d small = %d \n", x, small[x].x);
155 for(int i = E[x]; i; i = EDGE[i].nex) {
156 int y = EDGE[i].v;
157 getEXsmall(y, Node(small[x].x, small[x].d + EDGE[i].len));
158 }
159 return;
160 }
161
162 inline int getPos(int x, int f) {
163 int t = pw[d[x] - d[f]], y = x;
164 while(t >= 0) {
165 int mid = fa[y][t];
166 if(d[x] - d[mid] + small[x].d < d[mid] - d[f] + small[f].d) {
167 y = mid;
168 }
169 else if(d[x] - d[mid] + small[x].d == d[mid] - d[f] + small[f].d && small[f].x > small[x].x) {
170 y = mid;
171 }
172 t--;
173 }
174 return y;
175 }
176
177 void del(int x, int f) {
178 // printf("del x = %d small = %d %d \n", x, small[x].x, small[x].d);
179 if(f) {
180 if(small[x].x == small[f].x) {
181 SIZ[f] -= siz[x];
182 // printf("SIZ %d -= %d = %d \n", f, siz[x], SIZ[f]);
183 }
184 else {
185 int y = getPos(x, f);
186 SIZ[f] -= siz[y];
187 // printf("SIZ %d -= %d = %d \n", f, siz[y], SIZ[f]);
188 SIZ[x] = siz[y];
189 // printf("SIZ %d = siz %d %d \n", x, y, siz[y]);
190 }
191 }
192 for(int i = E[x]; i; i = EDGE[i].nex) {
193 int y = EDGE[i].v;
194 del(y, x);
195 }
196 ans[small[x].x] += SIZ[x];
197 return;
198 }
199
200 inline void solve() {
201 int k;
202 scanf("%d", &k);
203 Time++;
204 for(int i = 1; i <= k; i++) {
205 scanf("%d", &imp[i]);
206 now[imp[i]] = Time;
207 ans[imp[i]] = 0;
208 }
209 memcpy(imp2 + 1, imp + 1, k * sizeof(int));
210 build_t(k);
211
212 // out_t(RT);
213 // get Small
214 getSmall(RT); // get Size
215 getEXsmall(RT, Node(n + 1, INF));
216 //
217 SIZ[RT] = n;
218 del(RT, 0);
219 for(int i = 1; i <= k; i++) {
220 printf("%d ", ans[imp2[i]]);
221 }
222 printf("\n");
223 return;
224 }
225
226 int main() {
227
228 // freopen("in.in", "r", stdin);
229 // freopen("a.out", "w", stdout);
230
231 scanf("%d", &n);
232 for(int i = 1, x, y; i < n; i++) {
233 scanf("%d%d", &x, &y);
234 add(x, y);
235 add(y, x);
236 }
237 DFS_1(1, 0);
238 for(int i = 2; i <= n; i++) {
239 pw[i] = pw[i >> 1] + 1;
240 }
241 for(int j = 1; j <= pw[n]; j++) {
242 for(int i = 1; i <= n; i++) {
243 fa[i][j] = fa[fa[i][j - 1]][j - 1];
244 }
245 }
246 int q;
247 scanf("%d", &q);
248 while(q--) {
249 solve();
250 }
251 return 0;
252 }
AC代码
标签：洛谷,int,SIZ,世界,P3233,fa,EDGE,siz,small 来源： https://www.cnblogs.com/huyufeifei/p/10413471.html