求树上任意一点所能到达的最远距离 - 树上dp

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.

InputInput
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.OutputFor 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

题意 : 给你一颗树，以及树上两点之间的距离，求任意一点所能到的最远的距离。

思路分析：

　　对于这个问题，我们可以这样去思考，对于一颗树可以很方便的求出 以当前节点为根节点其所能到达的最远距离，但是这样并不能得到所有的节点的答案，其他的节点需要以其再考虑一下当前节点向上走的情况

再考虑 2 这个节点的时候其最优值可能来自 2 这个子树，也可能来自于右侧的这颗红色的树，即向上走

dp[x][0] 表示 x 节点向下走的最大值， dp[x][1]表示 x 节点向下走的次大值， dp[x][2] 表示 x 节点向上走的最大值。

代码示例：

const int maxn = 1e4+5;

int n;
struct node
{
    int to, cost;

    node(int _to=0, int _cost=0):to(_to), cost(_cost){}
};
vector<node>ve[maxn];
int dp[maxn][3];
int p[maxn];

void dfs1(int x, int fa){

    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i].to;
        int cost = ve[x][i].cost;
        if (to == fa) continue;

        dfs1(to, x);
        if (dp[x][0] <= dp[to][0]+cost){
            dp[x][1] = dp[x][0];
            dp[x][0] = dp[to][0]+cost;
            p[x] = to;
        }
        else if (dp[x][1] < dp[to][0]+cost){
            dp[x][1] = dp[to][0]+cost;
        }
    }
}
void dfs2(int x, int fa){

    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i].to;
        int cost = ve[x][i].cost;
        if (to == fa) continue;

        if (to != p[x]){
            dp[to][2] = max(dp[x][0]+cost, dp[x][2]+cost);
        }
        else dp[to][2] = max(dp[x][1]+cost, dp[x][2]+cost);
        dfs2(to, x);

        //printf("++++ %d %d  %d \n", x, to, dp[to][2]);
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int x, y;
    while(~scanf("%d", &n)){
        for(int i = 1; i <= n; i++) ve[i].clear();
        for(int i = 2; i <= n; i++){
            scanf("%d%d", &x, &y);
            ve[i].push_back(node(x, y));
            ve[x].push_back(node(i, y));
        }
        memset(dp, 0, sizeof(dp));
        dfs1(1, 0);
        dfs2(1, 0);

        for(int i = 1; i <= n; i++) {
            printf("%d\n", max(dp[i][0], dp[i][2]));
        }
    }

    return 0;
}

原文地址：https://www.cnblogs.com/ccut-ry/p/9188796.html