您的位置:首页 > Web前端 > Node.js

Light OJ 1257 Farthest Nodes in a Tree (II) (树形DP)

2016-02-15 11:41 696 查看
解析:树直径的模版题。

:

[code]#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
const int maxn = 30005;

struct Nod{
int b,val,next;
void init(int b,int val,int next){
this->b=b;this->val=val;this->next=next;
}
}buf[maxn<<1];
int n,m,len,E[maxn],dp[maxn][3],mlen[maxn];

void init(){
len = 0;
memset(E,-1,sizeof(E));
memset(dp,0,sizeof(dp));
memset(mlen,0,sizeof(mlen));
}
void addEdge(int a,int b,int val){
buf[len].init(b,val,E[a]);E[a]=len++;
buf[len].init(a,val,E[b]);E[b]=len++;
}
void dfs(int u,int pre){
int v,cost,i;
for(i = E[u];i != -1;i = buf[i].next){
v = buf[i].b;cost = buf[i].val;
if(v == pre) continue;
dfs(v,u);
if(dp[u][0] < dp[v][0] + cost){
dp[u][1] = dp[u][0];
dp[u][0] = dp[v][0]+cost;
}else if(dp[u][1] < dp[v][0]+cost) dp[u][1] = dp[v][0]+cost;
}
}
void ndfs(int u,int pre){
int dist,i;
for(i = E[u];i != -1;i = buf[i].next)
if(buf[i].b==pre){ dist = buf[i].val;break; }
if(pre != -1){
if(dp[u][0]+dist==dp[pre][0]){
dp[u][2] = dist + max(dp[pre][1],dp[pre][2]);
}else dp[u][2] = dist + max(dp[pre][0],dp[pre][2]);
}
for(i = E[u];i != -1;i = buf[i].next)
if(buf[i].b != pre) ndfs(buf[i].b,u);
}
int main(){
int i,j,cas,T,u,v,w;
scanf("%d",&cas);
for(T = 1;T <= cas;T++){
scanf("%d",&n);
init();
for(i = 1;i < n;i++){
scanf("%d%d%d",&u,&v,&w);
addEdge(u,v,w);
}
dfs(0,-1);ndfs(0,-1);
printf("Case %d:\n",T);
for(i = 0;i < n;i++){
mlen[i] = max(dp[i][0],dp[i][2]);
printf("%d\n",mlen[i]);
}
}

return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: