Housewife Wind
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 5471 | Accepted: 1371 |
Description
After their royal wedding, Jiajia and Wind hid
away in XX Village, to enjoy their ordinary happy life. People in XX Village
lived in beautiful huts. There are some pairs of huts connected by bidirectional
roads. We say that huts in the same pair directly connected. XX Village is so
special that we can reach any other huts starting from an arbitrary hut. If each
road cannot be walked along twice, then the route between every pair is
unique.
Since Jiajia earned enough money, Wind became a housewife.
Their children loved to go to other kids, then make a simple call to Wind:
‘Mummy, take me home!‘
At different times, the time needed to walk
along a road may be different. For example, Wind takes 5 minutes on a road
normally, but may take 10 minutes if there is a lovely little dog to play with,
or take 3 minutes if there is some unknown strange smell surrounding the
road.
Wind loves her children, so she would like to tell her
children the exact time she will spend on the roads. Can you help
her?
Input
The first line contains three integers n, q, s.
There are n huts in XX Village, q messages to process, and Wind is currently in
hut s. n < 100001 , q < 100001.
The following n-1 lines each
contains three integers a, b and w. That means there is a road directly
connecting hut a and b, time required is w. 1<=w<= 10000.
The
following q lines each is one of the following two types:
Message
A: 0 u
A kid in hut u calls Wind. She should go to hut u from her
current position.
Message B: 1 i w
The time required for i-th
road is changed to w. Note that the time change will not happen when Wind is on
her way. The changed can only happen when Wind is staying somewhere, waiting to
take the next kid.
Output
For each message A, print an integer X, the time
required to take the next child.
Sample Input
3 3 1
1 2 1
2 3 2
0 2
1 2 3
0 3
Sample Output
1
3
Source
POJ
Monthly--2006.02.26,zgl & twb
lca 加树状数组
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5
6 using namespace std;
7
8 const int MAX_N = 100005;
9 const int edge = MAX_N * 2;
10 int N,Q,S;
11 int first[MAX_N],Next[edge],w[edge],V[edge];
12 int id[MAX_N],vs[MAX_N * 2],dep[MAX_N * 2];
13 int d[2 * MAX_N][30],qid[2 * MAX_N][30];
14 int E[edge],c[MAX_N * 2];
15 int vis[MAX_N * 2];
16 int n;
17
18 int lowbit(int x) {
19 return x & (-x);
20 }
21
22 int sum(int x) {
23 int ret = 0;
24 while(x > 0) {
25 ret += c[x];
26 x -= lowbit(x);
27 }
28 return ret;
29 }
30
31 void add(int x,int d) {
32 //printf("x = %d\n",x);
33 while(x <= n) {
34 c[x] += d;
35 x += lowbit(x);
36 }
37 }
38
39 void add_edge(int id,int u) {
40 int e = first[u];
41 Next[id] = e;
42 first[u] = id;
43 }
44
45 int no(int x) {
46 if(x > N - 1) return x - N + 1;
47 else return x + N - 1;
48 }
49
50 void dfs(int u,int fa,int d,int &k,int m) {
51 id[u] = k;
52 vs[k] = u;
53 add(k,w[m]);
54 vis[m] = 1;
55 E[m] = k;
56 dep[k++] = d;
57 for(int e = first[u]; e != -1; e = Next[e]) {
58 if(V[e] != fa) {
59 dfs(V[e],u,d + 1,k,e);
60 vs[k] = u;
61 add(k,-w[e]);
62 vis[no(e)] = -1;
63 E[no(e)] = k;
64 dep[k++] = d;
65 }
66 }
67 }
68
69
70 void RMQ() {
71 for(int i = 1; i <= n; ++i) {
72 d[i][0] = dep[i];
73 qid[i][0] = i;
74 }
75 for(int j = 1; (1 << j) <= n; ++j) {
76 for(int i = 1; i + (1 << j) - 1 <= n; ++i) {
77 if(d[i][j - 1] > d[i + (1 << (j - 1))][j - 1]) {
78 d[i][j] = d[i + (1 << (j - 1))][j - 1];
79 qid[i][j] = qid[i + (1 << (j - 1))][j - 1];
80 } else {
81 d[i][j] = d[i][j - 1];
82 qid[i][j] = qid[i][j - 1];
83 }
84 }
85 }
86 }
87
88 int query(int L,int R) {
89 int k = 0;
90 //printf("L = %d R = %d\n",L,R);
91 while( (1 << (k + 1)) < (R - L + 1) ) ++k;
92 //printf("k = %d\n",k);
93 //printf("d= %d %d\n",d[L][1 << k] , d[R - (1 << k) + 1][1 << k]);
94 return d[L][k] < d[R - (1 << k) + 1][k] ?
95 qid[L][k] : qid[R - (1 << k) + 1][k];
96
97 }
98
99 int main()
100 {
101 // freopen("sw.in","r",stdin);
102 scanf("%d%d%d",&N,&Q,&S);
103 n = 2 * N - 1;
104 for(int i = 1; i <= N; ++i) first[i] = -1;
105 for(int i = 1; i <= N - 1; ++i) {
106 int u;
107 scanf("%d%d%d",&u,&V[i],&w[i]);
108 V[i + N - 1] = u;
109 w[i + N - 1] = w[i];
110 add_edge(i,u);
111 add_edge(i + N - 1,V[i]);
112 }
113
114 int k = 1;
115 dfs(S,-1,0,k,0);
116 RMQ();
117 //printf("k = %d\n",k);
118
119 int now = S;
120 for(int i = 1; i <= Q; ++i) {
121 int ch,v,id1;
122 scanf("%d",&ch);
123 if(ch == 0) {
124 scanf("%d",&v);
125 // printf("now = %d\n",now);
126 int p = vs[ query(min(id[now],id[v]),max(id[now],id[v])) ];
127 printf("%d\n",sum(id[now] ) + sum( id[v] ) - 2 * sum( id[p] ));
128 now = v;
129 } else {
130 scanf("%d%d",&id1,&v);
131 int d = v - w[id1];
132 w[id1] = v;
133 add(E[id1],vis[id1] * d);
134 add(E[id1 + N - 1],vis[id1 + N - 1] * d);
135 }
136 }
137
138
139 return 0;
140 }