Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem I

Roads in the North

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

 

Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are built in such a way that there is only one route from a village to a village that does not pass through some other village twice.

Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area.

The area has up to 10,000 villages connected by road segments. The villages are numbered from 1.

Input

The input contains several sets of input. Each set of input is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way. Two consecutive sets are separated by a blank line.

Output

For each set of input, you are to output a single line containing a single integer: the road distance between the two most remote villages in the area.

Sample Input

5 1 6

1 4 5

6 3 9

2 6 8

6 1 7

Sample Output

22

---

原本沒有權重的最長路徑，作法如下

做法 : 兩次 dfs,
1st dfs :
得到由下而上的最大高度 (要記錄第一高 第二高),
意即 得到每個節點的 subtree 的高度
2nd dfs :
將父節點高度轉過來, 這時要注意, 如果第一高是從這個節點接上去的, 那麼就抓第二高+1,

接著我們將 +1 的部分改成連接起來的 weight 值即可。

#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
struct ele {
    int to, v;
};
vector<ele> g[10001];
int dp1[10001][2], dp2[10001];
void init() {
    static int i;
    for(i = 1; i <= 10000; i++)
        g[i].clear(), dp1[i][0] = dp1[i][1] = dp2[i] = 0;
}
void dfs1(int nd, int p) {
    for(vector<ele>::iterator it = g[nd].begin();
        it != g[nd].end(); it++) {
        if(it->to != p) {
            dfs1(it->to, nd);
            if(dp1[it->to][0] + it->v > dp1[nd][1])
                dp1[nd][1] = dp1[it->to][0] + it->v;
            if(dp1[nd][1] > dp1[nd][0])
                swap(dp1[nd][0], dp1[nd][1]);
        }
    }
}
void dfs2(int nd, int p, int st, int w) {
    if(nd != st) {
        if(dp1[nd][0] + w == dp1[p][0])
            dp2[nd] = max(dp2[p], dp1[p][1])+w;
        else
            dp2[nd] = max(dp2[p], dp1[p][0])+w;
    }
    for(vector<ele>::iterator it = g[nd].begin();
        it != g[nd].end(); it++) {
        if(it->to != p) {
            dfs2(it->to, nd, st, it->v);
        }
    }
}
void sol(int st) {
    dfs1(st, -1);
    dfs2(st, -1, st, 0);
    int ans = 0, i;
    for(i = 1; i <= 10000; i++) {
        ans = max(ans, dp1[i][0]+dp1[i][1]);
        ans = max(ans, dp1[i][0]+dp2[i]);
    }
    printf("%d\n", ans);
}
int main() {
    char cmd[100];
    ele E;
    int x, y, v;
    while(gets(cmd)) {
        if(cmd[0] == '\0') {
            sol(x);
            init();
            continue;
        }
        sscanf(cmd, "%d %d %d", &x, &y, &v);
        E.to = x, E.v = v;
        g[y].push_back(E);
        E.to = y;
        g[x].push_back(E);
    }
    sol(x);
    return 0;
}