lght oj 1257 - Farthest Nodes in a Tree (II) (树dp)
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题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1257
跟hdu2196一样,两次dfs
1 //#pragma comment(linker, "/STACK:102400000, 102400000") 2 #include <algorithm> 3 #include <iostream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <vector> 8 #include <cmath> 9 #include <ctime> 10 #include <list> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL; 15 typedef pair <int, int> P; 16 const int N = 1e5 + 5; 17 struct Edge { 18 int next, to, cost; 19 }edge[N << 1]; 20 int head[N], tot; 21 P d[N], pos[N]; 22 int up[N]; 23 24 void init(int n) { 25 memset(head, -1, sizeof(head)); 26 tot = 0; 27 for(int i = 0; i <= n; ++i) { 28 d[i].first = d[i].second = 0; 29 pos[i].first = pos[i].second = -1; 30 up[i] = 0; 31 } 32 } 33 34 inline void add(int u, int v, int cost) { 35 edge[tot].next = head[u]; 36 edge[tot].to = v; 37 edge[tot].cost = cost; 38 head[u] = tot++; 39 } 40 41 void dfs1(int u, int p) { 42 for(int i = head[u]; ~i; i = edge[i].next) { 43 int v = edge[i].to; 44 if(v == p) 45 continue; 46 dfs1(v, u); 47 if(d[v].first + edge[i].cost > d[u].first) { 48 if(d[u].first != 0) 49 d[u].second = d[u].first; 50 d[u].first = d[v].first + edge[i].cost; 51 pos[u].first = v; 52 } else if(d[v].first + edge[i].cost > d[u].second) { 53 d[u].second = d[v].first + edge[i].cost; 54 pos[u].second = v; 55 } 56 } 57 } 58 59 void dfs2(int u, int p) { 60 for(int i = head[u]; ~i; i = edge[i].next) { 61 int v = edge[i].to; 62 if(v == p) 63 continue; 64 if(v == pos[u].first) { 65 up[v] = max(up[u], d[u].second) + edge[i].cost; 66 } else { 67 up[v] = max(up[u], d[u].first) + edge[i].cost; 68 } 69 dfs2(v, u); 70 } 71 } 72 73 int main() 74 { 75 int t, n; 76 scanf("%d", &t); 77 for(int ca = 1; ca <= t; ++ca) { 78 scanf("%d", &n); 79 init(n); 80 int u, v, cost; 81 for(int i = 1; i < n; ++i) { 82 scanf("%d %d %d", &u, &v, &cost); 83 add(u, v, cost); 84 add(v, u, cost); 85 } 86 dfs1(0, -1); 87 dfs2(0, -1); 88 printf("Case %d:\n", ca); 89 for(int i = 0; i < n; ++i) { 90 printf("%d\n", max(d[i].first, up[i])); 91 } 92 } 93 return 0; 94 }
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