Connections between cities
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5037 Accepted Submission(s): 1399Problem Description
After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.Input
Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.
Output
For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.
Sample Input
5 3 2
1 3 2
2 4 3
5 2 3
1 4
4 5Sample Output
Not connected
6Hint
Hint
Huge input, scanf recommended.
给出一个森林，然后有Ｑ组询问，每次询问给出ｕ，ｖ，问ｕ和ｖ是否连通？如果连通则输出ｕ到ｖ的最短距离。
由于给出的 并不是一棵树，所以可以加个根节点，但是开始的时候我是从０出发到每个结点都加了一条边，结果TLE.
后来加了并查集，然后从０出发到每个连通分量只加一条边。。轻松水过。
Accepted Code:
1 /*************************************************************************
2 > File Name: 2874.cpp
3 > Author: Stomach_ache
4 > Mail: [email protected]
5 > Created Time: 2014年09月24日 星期三 20时14分04秒
6 > Propose:
7 ************************************************************************/
8 #include <cmath>
9 #include <string>
10 #include <cstdio>
11 #include <vector>
12 #include <fstream>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17 /*Let‘s fight!!!*/
18
19 const int MAX_N = 10050;
20 const int MAX_LOG = 15;
21 typedef pair<int, int> pii;
22 int N, M, Q;
23 int p[MAX_N][MAX_LOG], depth[MAX_N], Dist[MAX_N], fa[MAX_N];
24 vector<pii> G[MAX_N];
25
26 void dfs(int u, int fa, int d, int cost) {
27 p[u][0] = fa;
28 depth[u] = d;
29 Dist[u] = Dist[fa] + cost;
30 for (int i = 0; i < G[u].size(); i++) {
31 int v = G[u][i].first;
32 if (v != fa) dfs(v, u, d + 1, G[u][i].second);
33 }
34 }
35
36 void init() {
37 dfs(0, -1, 0, 0);
38
39 for (int k = 0; k + 1 < MAX_LOG; k++) {
40 for (int v = 0; v <= N; v++) {
41 if (p[v][k] < 0) p[v][k + 1] = -1;
42 else p[v][k + 1] = p[p[v][k]][k];
43 }
44 }
45 }
46
47 int lca(int u, int v) {
48 if (depth[u] > depth[v]) swap(u, v);
49 for (int k = 0; k < MAX_LOG; k++) {
50 if ((depth[v] - depth[u]) >> k & 1) {
51 v = p[v][k];
52 }
53 }
54 if (u == v) return u;
55 for (int k = MAX_LOG - 1; k >= 0; k--) {
56 if (p[u][k] != p[v][k]) {
57 u = p[u][k];
58 v = p[v][k];
59 }
60 }
61 return p[u][0];
62 } 68
69 int findfa(int x) {
70 return x != fa[x] ? fa[x] = findfa(fa[x]) : x;
71 }
72
73 void read(int &res) {
74 res = 0;
75 char c = ‘ ‘;
76 while (c < ‘0‘ || c > ‘9‘) c = getchar();
77 while (c >= ‘0‘ && c <= ‘9‘) res = res * 10 + c - ‘0‘, c = getchar();
78 }
79
80 int main(void) {
81 while (~scanf("%d %d %d", &N, &M, &Q)) {
82 for (int i = 0; i <= N; i++) G[i].clear(), fa[i] = i;
83
84 for (int i = 1; i <= M; i++) {
85 int u, v, w;
86 read(u), read(v), read(w);
87 //scanf("%d %d %d", &u, &v, &w);
88 G[u].push_back(pii(v, w));
89 G[v].push_back(pii(u, w));
90 int x = findfa(u), y = findfa(v);
91 if (x != y) {
92 fa[y] = x;
93 }
94 }
95 for (int i = 1; i <= N; i++) if (fa[i] == i) {
96 G[0].push_back(pii(i, 0));
97 G[i].push_back(pii(0, 0));
98 }
99 init();
100
101 while (Q--) {
102 int u, v;
103 read(u), read(v);
104 //scanf("%d %d", &u, &v);
105 int x = lca(u, v);
106 if (x == 0) puts("Not connected");
107 else printf("%d\n", Dist[u] + Dist[v] - 2 * Dist[x]);
108 }
109 }
110 return 0;
111 }