「hdu2196」Computer
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.
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
5 1 1 2 1 3 1 1 1
Sample Output
3 2 3 4 4
题解
树形dp。。。f[x]和g[x]分别表示从x出发，终点在/不在x的子树的最长距离
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#include<iostream> #include<cstdio> #define ll long long using namespace std; #define inf 1000000000 inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,cnt,T; int ans[10005],last[10005],f[10005],g[10005]; struct edge{int to,next,v;}e[20005]; void insert(int u,int v,int w) { e[++cnt].to=v;e[cnt].next=last[u];e[cnt].v=w;last[u]=cnt; e[++cnt].to=u;e[cnt].next=last[v];e[cnt].v=w;last[v]=cnt; } void dp(int x,int fa) { for(int i=last[x];i;i=e[i].next) if(e[i].to!=fa) { dp(e[i].to,x); f[x]=max(f[x],f[e[i].to]+e[i].v); } } void solve(int x,int fa) { int k=0,t=0,v1=0,v2=0; for(int i=last[x];i;i=e[i].next) if(e[i].to!=fa) if(f[e[i].to]+e[i].v>=f[k]+v1)k=e[i].to,v1=e[i].v; for(int i=last[x];i;i=e[i].next) if(e[i].to!=fa&&e[i].to!=k) { g[e[i].to]=ans[e[i].to]=f[k]+e[i].v+v1; if(f[e[i].to]+e[i].v>=f[t]+v2)t=e[i].to,v2=e[i].v; } g[k]=ans[k]=f[t]+v1+v2; for(int i=last[x];i;i=e[i].next) if(e[i].to!=fa) g[e[i].to]=max(g[e[i].to],g[x]+e[i].v); for(int i=last[x];i;i=e[i].next) if(e[i].to!=fa) { ans[e[i].to]=max(ans[e[i].to],g[x]+e[i].v); solve(e[i].to,x); } } int main() { while(scanf("%d",&n)!=EOF) { cnt=0; memset(last,0,sizeof(last)); memset(f,0,sizeof(f)); memset(g,0,sizeof(g)); for(int i=2;i<=n;i++) { int u=read(),v=read(); insert(u,i,v); } f[0]=g[0]=-inf; dp(1,0); solve(1,0); for(int i=1;i<=n;i++) printf("%d\n",max(f[i],ans[i])); } return 0; } |
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