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POJ 1986 Distance Queries
http://poj.org/problem?id=1986
题意:一棵树里找到两个点的距离。(不用考虑不联通的情况)
题解:LCA模板题。
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 #include <string> 5 #include <cstdio> 6 #include <cmath> 7 #include <queue> 8 #include <map> 9 #include <set> 10 11 #define eps 1e-5 12 #define MAXN 55555 13 #define MAXM 111111 14 #define INF 1000000000 15 using namespace std; 16 int n, m, q; 17 struct edge 18 { 19 int v, next, w; 20 }es[MAXM]; 21 int head[MAXN], e; 22 int ide,tmpdfn; 23 int f[2 * MAXN], id[MAXN], used[MAXN], pos[MAXN], dis[MAXN]; 24 int mi[2 * MAXN][18]; 25 26 void init() 27 { 28 memset(head, -1, sizeof(head)); 29 e = 0; 30 ide = tmpdfn = 0; 31 memset(used, 0, sizeof(used)); 32 dis[1] = 0; 33 } 34 35 void add(int u, int v, int w) 36 { 37 es[e].v = v; 38 es[e].w = w; 39 es[e].next = head[u]; 40 head[u] = e++; 41 } 42 void dfs(int u) 43 { 44 used[u] = 1; 45 int tmp = ++tmpdfn; 46 f[++ide] = tmp; 47 id[tmp] = u; 48 pos[u] = ide; 49 for(int i = head[u]; i != -1; i = es[i].next) 50 { 51 int v = es[i].v; 52 if(!used[v]) 53 { 54 dis[v] = dis[u] + es[i].w; 55 dfs(v); 56 f[++ide] = tmp; 57 } 58 } 59 60 } 61 void rmqinit(int n, int *w) 62 { 63 for(int i = 1; i <= n; i++) mi[i][0] = w[i]; 64 int m = (int)(log(n * 1.0) / log(2.0)); 65 for(int i = 1; i <= m; i++) 66 for(int j = 1; j <= n; j++) 67 { 68 mi[j][i] = mi[j][i - 1]; 69 if(j + (1 << (i - 1)) <= n) mi[j][i] = min(mi[j][i], mi[j + (1 << (i - 1))][i - 1]); 70 } 71 } 72 73 int rmqmin(int l,int r) 74 { 75 int m = (int)(log((r - l + 1) * 1.0) / log(2.0)); 76 return min(mi[l][m] , mi[r - (1 << m) + 1][m]); 77 } 78 79 int LCA(int l, int r) 80 { 81 if(pos[l] > pos[r]) swap(l, r); 82 int ans = rmqmin(pos[l], pos[r]); 83 return id[ans]; 84 } 85 int main() 86 { 87 // freopen("/Users/apple/Desktop/题/POJ 1986_1/POJ 1986_1/in","r",stdin); 88 scanf("%d%d", &n, &m); 89 int u, v, w, l, r; 90 init(); 91 while(m--) 92 { 93 scanf("%d%d%d%*s", &u, &v, &w); 94 add(u, v, w); 95 add(v, u, w); 96 } 97 dfs(1); 98 rmqinit(ide, f); 99 scanf("%d", &q);100 while(q--)101 {102 scanf("%d%d", &l, &r);103 printf("%d\n", dis[l] + dis[r] - 2 * dis[LCA(l, r)]);104 }105 return 0;106 }
POJ 1986 Distance Queries
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