Problem 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.
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 5
Sample Output
Not connected 6
Hint
Hint Huge input, scanf recommended.
题意是说给你一个森林，让你求两点之间的最近距离。
lca求最近公共祖先，如果不是在同一棵树上，则输出Not connected。
用并查集来判断是否在同一颗树上面
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <cmath> 5 #include <algorithm> 6 #include <set> 7 #include <iostream> 8 #include <map> 9 #include <stack> 10 #include <string> 11 #include <vector> 12 #define pi acos(-1.0) 13 #define eps 1e-6 14 #define fi first 15 #define se second 16 #define lson l,m,rt<<1 17 #define rson m+1,r,rt<<1|1 18 #define bug printf("******\n") 19 #define mem(a,b) memset(a,b,sizeof(a)) 20 #define fuck(x) cout<<"["<<x<<"]"<<endl 21 #define f(a) a*a 22 #define sf(n) scanf("%d", &n) 23 #define sff(a,b) scanf("%d %d", &a, &b) 24 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) 25 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d) 26 #define pf printf 27 #define FRE(i,a,b) for(i = a; i <= b; i++) 28 #define FREE(i,a,b) for(i = a; i >= b; i--) 29 #define FRL(i,a,b) for(i = a; i < b; i++) 30 #define FRLL(i,a,b) for(i = a; i > b; i--) 31 #define FIN freopen("DATA.txt","r",stdin) 32 #define gcd(a,b) __gcd(a,b) 33 #define lowbit(x) x&-x 34 #pragma comment (linker,"/STACK:102400000,102400000") 35 using namespace std; 36 typedef long long LL; 37 typedef unsigned long long ULL; 38 const int maxn = 1e5 + 10; 39 int _pow[maxn], dep[maxn], dis[maxn], vis[maxn], ver[maxn]; 40 int tot, head[maxn], dp[maxn * 2][25], k, first[maxn], fa[maxn]; 41 struct node { 42 int u, v, w, nxt; 43 } edge[maxn << 2]; 44 void init() { 45 tot = 0; 46 mem(head, -1); 47 for (int i = 0 ; i < maxn ; i++) fa[i] = i; 48 } 49 int Find(int x) { 50 return x == fa[x] ? fa[x] : fa[x] = Find(fa[x]); 51 } 52 void combine(int x, int y) { 53 int nx = Find(x), ny = Find(y); 54 if(nx != ny) fa[nx] = ny; 55 return ; 56 } 57 void add(int u, int v, int w) { 58 edge[tot].v = v, edge[tot].u = u; 59 edge[tot].w = w, edge[tot].nxt = head[u]; 60 head[u] = tot++; 61 } 62 void dfs(int u, int DEP) { 63 vis[u] = 1; 64 ver[++k] = u; 65 first[u] = k; 66 dep[k] = DEP; 67 for (int i = head[u]; ~i; i = edge[i].nxt) { 68 if (vis[edge[i].v]) continue; 69 int v = edge[i].v, w = edge[i].w; 70 dis[v] = dis[u] + w; 71 dfs(v, DEP + 1); 72 ver[++k] = u; 73 dep[k] = DEP; 74 } 75 } 76 void ST(int len) { 77 int K = (int)(log((double)len) / log(2.0)); 78 for (int i = 1 ; i <= len ; i++) dp[i][0] = i; 79 for (int j = 1 ; j <= K ; j++) { 80 for (int i = 1 ; i + _pow[j] - 1 <= len ; i++) { 81 int a = dp[i][j - 1], b = dp[i + _pow[j - 1]][j - 1]; 82 if (dep[a] < dep[b]) dp[i][j] = a; 83 else dp[i][j] = b; 84 } 85 } 86 } 87 int RMQ(int x, int y) { 88 int K = (int)(log((double)(y - x + 1)) / log(2.0)); 89 int a = dp[x][K], b = dp[y - _pow[K] + 1][K]; 90 if (dep[a] < dep[b]) return a; 91 else return b; 92 } 93 int LCA(int u, int v) { 94 int x = first[u], y = first[v]; 95 if (x > y) swap(x, y); 96 int ret = RMQ(x, y); 97 return ver[ret]; 98 } 99 int main() { 100 for (int i = 0 ; i < 40 ; i++) _pow[i] = (1 << i); 101 int n, m, q; 102 while(~sfff(n, m, q)) { 103 init(); 104 mem(vis, 0); 105 for (int i = 0 ; i < m ; i++) { 106 int u, v, w; 107 sfff(u, v, w); 108 add(u, v, w); 109 add(v, u, w); 110 combine(u, v); 111 } 112 k = 0; 113 for (int i = 1 ; i <= n ; i++) { 114 if (fa[i] == i) { 115 dis[i] = 0; 116 dfs(i, 1); 117 } 118 } 119 ST(2 * n - 1); 120 while(q--) { 121 int u, v; 122 sff(u, v); 123 int lca = LCA(u, v); 124 if (Find(u) == Find(v)) printf("%d\n", dis[u] + dis[v] - 2 * dis[lca]); 125 else printf("Not connected\n"); 126 } 127 } 128 return 0; 129 }