Mart Master II
Time Limit: 6000ms
Memory Limit: 65536KB
This problem will be judged on HDU. Original ID: 5016
64-bit integer IO format: %I64d Java class name: Main
Trader Dogy lives in city S, which consists of n districts. There are n - 1 bidirectional roads in city S, each connects a pair of districts. Indeed, city S is connected, i.e. people can travel between every pair of districts by roads.
In some districts there are marts founded by Dogy’s competitors. when people go to marts, they’ll choose the nearest one. In cases there are more than one nearest marts, they’ll choose the one with minimal city number.
Dogy’s money could support him to build only one new marts, he wants to attract as many people as possible, that is, to build his marts in some way that maximize the number of people who will choose his mart as favorite. Could you help him?
Input
There are multiple test cases. Please process till EOF.
In each test case:
First line: an integer n indicating the number of districts.
Next n - 1 lines: each contains three numbers bi, ei and wi, (1 ≤ bi,ei ≤ n,1 ≤ wi ≤ 10000), indicates that there’s one road connecting city bi and ei, and its length is wi.
Last line : n(1 ≤ n ≤ 105) numbers, each number is either 0 or 1, i-th number is 1 indicates that the i-th district has mart in the beginning and vice versa.
Output
For each test case, output one number, denotes the number of people you can attract, taking district as a unit.
Sample Input
5 1 2 1 2 3 1 3 4 1 4 5 1 1 0 0 0 1 5 1 2 1 2 3 1 3 4 1 4 5 1 1 0 0 0 0 1 1 1 0
Sample Output
2 4 0 1 解题：挺恶心的一道题
1 #include <bits/stdc++.h>
2 using namespace std;
3 using PII = pair<int,int>;
4 const int maxn = 100010;
5 const int INF = ~0u>>2;
6 PII d[3][maxn];
7 struct arc {
8 int to,w,next;
9 arc(int x = 0,int y = 0,int z = -1) {
10 to = x;
11 w = y;
12 next = z;
13 }
14 } e[maxn<<1];
15 int head[maxn],sz[maxn],maxson[maxn],mart[maxn],tot,cnt,n;
16 int ans[maxn];
17 void add(int u,int v,int w) {
18 e[tot] = arc(v,w,head[u]);
19 head[u] = tot++;
20 }
21 queue<int>q;
22 bool done[maxn];
23 void spfa() {
24 memset(done,false,sizeof done);
25 while(!q.empty()){
26 int u = q.front();
27 q.pop();
28 done[u] = false;
29 for(int i = head[u]; ~i; i = e[i].next){
30 PII tmp(d[0][u].first + e[i].w,d[0][u].second);
31 if(d[0][e[i].to] > tmp){
32 d[0][e[i].to] = tmp;
33 if(!done[e[i].to]){
34 done[e[i].to] = true;
35 q.push(e[i].to);
36 }
37 }
38 }
39 }
40 }
41 int dfs(int u,int fa){
42 sz[u] = 1;
43 maxson[u] = 0;
44 for(int i = head[u]; ~i; i = e[i].next){
45 if(e[i].to == fa || done[e[i].to]) continue;
46 dfs(e[i].to,u);
47 sz[u] += sz[e[i].to];
48 maxson[u] = max(maxson[u],sz[e[i].to]);
49 }
50 return sz[u];
51 }
52 int root(const int sum,int u,int fa){
53 int ret = u;
54 maxson[u] = max(maxson[u],sum - sz[u]);
55 for(int i = head[u]; ~i; i = e[i].next){
56 if(e[i].to == fa || done[e[i].to]) continue;
57 int x = root(sum,e[i].to,u);
58 if(maxson[x] < maxson[ret]) ret = x;
59 }
60 return ret;
61 }
62 void update(int u,int w,int fa){
63 d[1][cnt] = PII(w,u);
64 d[2][cnt++] = PII(d[0][u].first - w,d[0][u].second);
65 for(int i = head[u]; ~i; i = e[i].next){
66 if(e[i].to == fa || done[e[i].to]) continue;
67 update(e[i].to,w + e[i].w,u);
68 }
69 }
70 void calc(int u,int w,int sg){
71 cnt = 0;
72 update(u,w,0);
73 sort(d[2],d[2] + cnt);
74 for(int i = 0; i < cnt; ++i){
75 if(mart[d[1][i].second]) continue;
76 auto it = lower_bound(d[2],d[2] + cnt,d[1][i]) - d[2];
77 ans[d[1][i].second] += (cnt - it)*sg;
78 }
79 }
80 void solve(int u){
81 int rt = root(dfs(u,0),u,0);
82 done[rt] = true;
83 calc(rt,0,1);
84 for(int i = head[rt]; ~i; i = e[i].next){
85 if(done[e[i].to]) continue;
86 calc(e[i].to,e[i].w,-1);
87 }
88 for(int i = head[rt]; ~i; i = e[i].next){
89 if(done[e[i].to]) continue;
90 solve(e[i].to);
91 }
92 }
93 int main() {
94 int u,v,w;
95 while(~scanf("%d",&n)) {
96 memset(head,-1,sizeof head);
97 int ret = tot = 0;
98 for(int i = 1; i < n; ++i) {
99 scanf("%d%d%d",&u,&v,&w);
100 add(u,v,w);
101 add(v,u,w);
102 }
103 for(int i = 1; i <= n; ++i) {
104 scanf("%d",mart + i);
105 if(mart[i]) {
106 d[0][i] = PII(0,i);
107 q.push(i);
108 } else d[0][i] = PII(INF,0);
109 ans[i] = 0;
110 }
111 spfa();
112 solve(1);
113 for(int i = 1; i <= n; ++i)
114 ret = max(ret,ans[i]);
115 printf("%d\n",ret);
116 }
117 return 0;
118 }