HDU P2196 Computer

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题目

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

51 12 13 11 1

Sample Output

32344

Author
scnu

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题目大意

给出一个数 n，表示树的大小，然后给出 n−1 条关系，其中第 i 条为 x,y 表示 i 与 x 之间有一条长为 y 的边（该树的根节点为1）

然后，算出树中各点到离其最远的点的距离

多组数据

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题解

树形DP

f[i][0] 表示以i为根的子树时，i 到离它最远的点的距离

f[i][1] 表示以 fa[i] 为根的树删去以 i 为根的子树时，fa[i] 到离它最远的点的距离

显然，对于节点 i 离它最远的点的距离即为 max(f[i][0],f[i][1]+e[i,fa[i]])

然后，就是转移方程了

对于 f[i][0] 有 f[i][0]=max(f[i][0],f[son[i]][0]+e[i,son[i]])

对于 f[i][1] 有 f[i][1]=max(f[i][1],t[fa[i]]+e[fa[i],i])，其中 t[fa[i]] 为不经过 i 的离 fa[i] 最长距离，即从 f[son[fa[i]]][0],f[fa[i]][1] 中挑出两个后判断是否经过 i 就可以了

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代码

#include<cstdio>#include<cstring>using namespace std;int n,tot;int lnk[10005],f[10005][2];struct edge{    int nxt,y,v;} e[20005];int readln(){    int x=0;    char ch=getchar();    while (ch<'0'||ch>'9') ch=getchar();    while ('0'<=ch&&ch<='9') x=x*10+ch-48,ch=getchar();    return x;}void add(int x,int y,int v){    tot++;e[tot].nxt=lnk[x];lnk[x]=tot;e[tot].y=y;e[tot].v=v;}int max(int x,int y){return x>y?x:y;}void dfs1(int x,int fa){    int v,y;    for (int i=lnk[x];i;i=e[i].nxt)    {        y=e[i].y,v=e[i].v;        if (fa==y) continue;        dfs1(y,x);        f[x][0]=max(f[x][0],f[y][0]+v);    }}void dfs2(int x,int fa){    int t0=0,t1=0,p0=0,p1=0,v,y;    for (int i=lnk[x];i;i=e[i].nxt)    {        y=e[i].y,v=e[i].v;        if (fa==y) continue;        if (f[y][0]+v>t0) t1=t0,t0=f[y][0]+v,p1=p0,p0=y;        else if (f[y][0]+v>t1) t1=f[y][0]+v,p1=y;    }    for (int i=lnk[x];i;i=e[i].nxt)    {        y=e[i].y,v=e[i].v;        if (fa==y) continue;        f[y][1]=f[x][1]+v;        if (p0!=y) {            if (p0!=0) f[y][1]=max(f[y][1],t0+v);        } else        if (p1!=y)        {            if (p1!=0) f[y][1]=max(f[y][1],t1+v);        }        dfs2(y,x);    }}int main(){    while (~scanf("%d",&n))    {        memset(f,0,sizeof(f));        memset(lnk,0,sizeof(lnk));        tot=0;        int x,y;        for (int i=2;i<=n;i++)        {            x=readln(),y=readln();            add(i,x,y);add(x,i,y);        }        dfs1(1,0);        dfs2(1,0);        for (int i=1;i<=n;i++) printf("%d\n",max(f[i][0],f[i][1]));    }    return 0;}