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任意門：http://poj.org/problem?id=1986

Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 16648		Accepted: 5817
Case Time Limit: 1000MS

Description

Farmer John‘s cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ‘s distance queries as quickly as possible!

Input

* Lines 1..1+M: Same format as "Navigation Nightmare"

* Line 2+M: A single integer, K. 1 <= K <= 10,000

* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.

Output

* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.

Sample Input

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6
1 4
2 6

Sample Output

13
3
36

Hint

Farms 2 and 6 are 20+3+13=36 apart.

題意概括：

輸入：

第一行輸入結點數 N 和 邊數 M。

接下來 M 行輸入邊的信息：起點 u 終點 v 距離 w 方向 s

輸入查詢數 K

接下來 K 行 輸入查詢值： 起點 u，終點 v；

解題思路：

一個有向無環圖，當作一棵樹來處理，根結點隨意，假設為 1；

LCA 兩點最短距離 老套路。

AC code：

 1
 
 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <cstring>
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 using namespace std;
 9 const int MAXN = 5e5+5;
10 struct Edge{int v, w, nxt;}edge[MAXN<<1];
11 struct Query
12 {
13     int v, id;
14     Query(){};
15     Query(int _v, int _id):v(_v),id(_id){};
16 };
17 vector<Query> q[MAXN];
18 
19 int head[MAXN], cnt;
20 int dis[MAXN];
21 int fa[MAXN];
22 bool vis[MAXN];
23 int ans[MAXN];
24 int N, M, K;
25 
26 void init()
27 {
28     memset(vis, false, sizeof(vis));
29     memset(head, -1, sizeof(head));
30     memset(dis, 0, sizeof(dis));
31     memset(ans, 0, sizeof(ans));
32     for(int i = 0; i <= N; i++) q[i].clear();
33     cnt = 0;
34 }
35 
36 int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
37 
38 void AddEdge(int from, int to, int weight)
39 {
40     edge[cnt].v = to;
41     edge[cnt].w = weight;
42     edge[cnt].nxt = head[from];
43     head[from] = cnt++;
44 }
45 
46 void dfs(int s, int f)
47 {
48     int root = s;
49     for(int i = head[s]; i != -1; i = edge[i].nxt){
50         if(edge[i].v == f) continue;
51         dis[edge[i].v] = dis[root] + edge[i].w;
52         dfs(edge[i].v, s);
53     }
54 }
55 
56 void Tarjan(int s, int f)
57 {
58     int root = s;
59     fa[s] = s;
60     for(int i = head[s]; i != -1; i = edge[i].nxt){
61         int Eiv = edge[i].v;
62         if(Eiv == f) continue;
63         Tarjan(Eiv, root);
64         fa[getfa(Eiv)] = root;
65     }
66     vis[s] = true;
67     for(int i = 0; i < q[s].size(); i++){
68         if(vis[q[s][i].v] && ans[q[s][i].id] == 0){
69             ans[q[s][i].id] = dis[q[s][i].v] + dis[s] - 2*dis[getfa(q[s][i].v)];
70         }
71     }
72 }
73 
74 int main()
75 {
76     scanf("%d%d", &N, &M);
77     init();
78     char s;
79     for(int i = 1, u, v, w; i <= M; i++){
80         scanf("%d%d%d %c", &u, &v, &w, &s);
81         AddEdge(u, v, w);
82         AddEdge(v, u, w);
83     }
84     scanf("%d", &K);
85     for(int i = 1, u, v; i <= K; i++){
86         scanf("%d%d", &u, &v);
87         q[u].push_back(Query(v, i));
88         q[v].push_back(Query(u, i));
89     }
90     dfs(1, -1);
91     Tarjan(1, -1);
92     for(int i = 1; i <= K; i++){
93         printf("%d\n", ans[i]);
94     }
95     return 0;
96 }

View Code

POJ 1986 Distance Queries 【輸入YY && LCA（Tarjan離線）】