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hdu2196 Computer（樹形dp）

編輯：C++入門知識

hdu2196 Computer（樹形dp）

Computer

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4715 Accepted Submission(s): 2376

Problem Description A school bought the first computer some time ago(so this computer's id is 1). During the recent years the school bought N-1 new computers. Each new computer was connected to one of settled earlier. Managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which i-th computer needs to send signal (i.e. length of cable to the most distant computer). You need to provide this information.

Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.
Input Input file contains multiple test cases.In each case there is natural number N (N<=10000) in the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two natural numbers - number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input are separated by a space.
Output For each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N).
Sample Input

Sample Output

3
2
3
4
4
題意：給一棵樹，每條樹邊都有權值，問從每個頂點出發，經過的路徑權值之和最大為多少？每條樹邊都只能走一次。分析：以每個結點為根找到最長的子節點鏈，加上該結點為孩子的另一支子樹鏈。詳細解析

#include  #include  #include  #include  #include  #include#include  #include  #include  #include  using namespace std; const double eps = 1e-6; const double pi = acos(-1.0); const int INF = 0x3f3f3f3f; const int MOD = 1000000007; #define ll long long #define CL(a,b) memset(a,b,sizeof(a)) #define MAXN 10005 struct node { int v,len,sum; node *next; }tree[MAXN*2], *head[MAXN*2]; int n,ptr,vis[MAXN]; int dp[MAXN]; void init() { ptr=1; CL(vis, 0); CL(dp, 0); CL(head, NULL); } void add(int s, int e, int len)//建樹 { tree[ptr].v = e; tree[ptr].len = len; tree[ptr].next = head[s];head[s] = &tree[ptr++]; tree[ptr].v = s; tree[ptr].len = len; tree[ptr].next = head[e],head[e] = &tree[ptr++]; } void dfs1(int idx)//以idx為根的最長鏈 { vis[idx] = 1; node *p = head[idx]; while(p != NULL) { if(!vis[p->v]) { dfs1(p->v); dp[idx] = max(dp[idx], dp[p->v]+p->len); p->sum = dp[p->v]+p->len; } p = p->next; } } void dfs2(int father, int child)//同根的另一支子樹 { if(vis[child]) return ; vis[child] = 1; int maxx = 0; node *p = head[father]; while(p != NULL)//找到父節點除child外其他分支的最大價值 { if(p->v != child) maxx = max(maxx, p->sum); p = p->next; } p = head[child]; while(p !=NULL)//繼續往上更新，才能保證每步都得到最優解 { if(p->v == father) { p->sum = p->len + maxx; break; } p = p->next; } p = head[child]; while(p != NULL)//每次都更新當前節點，並往下遞歸計算,父節點會因為vis=1而不計算 { dp[child] = max(dp[child], p->sum); dfs2(child, p->v); p = p->next; } } int main() { int a,b; while(scanf(%d,&n)==1) { init(); for(int i=2; i<=n; i++) { scanf(%d%d,&a,&b); add(i, a, b); } dfs1(1); CL(vis, 0); node *p = head[1]; while(p != NULL) { dfs2(1, p->v); p = p->next; } for(int i=1; i<=n; i++) printf(%d ,dp[i]); } return 0; }