POJ 1985 Cow Marathon （树形DP，树的直径）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>

using namespace std;
const int maxn = 1e5 + 5;
int d[maxn];
bool vis[maxn], vvis[maxn];//vis是BFS的标记，vvis是总的标记
vector<int> G[maxn], w[maxn];
//思路就是先在树上找一个点，然后从这个点开始找，找最远的点，
//然后两从最远的点找最远的点，这个所有点中的最大值就是树的直径
//d[i]表示到结点 i 的最长距离
int bfs(int root){
memset(vis, false, sizeof(vis));
memset(d, 0, sizeof(d));
queue<int> q;
q.push(root);
int ans = root, m = 0;
vis[root] = vvis[root] = true;
while(!q.empty()){
root = q.front();   q.pop();
for(int i = 0; i < G[root].size(); ++i){
int u = G[root][i];
if(vis[u])  continue;
q.push(u);
vis[u] = vvis[u] = true;
d[u] = d[root] + w[root][i];
if(d[u] > m){
ans = u;
m = d[u];
}
}
}
return ans;
}

void init(int n){
for(int i = 1; i <= n; ++i){
G[i].clear();
w[i].clear();
vvis[i] = false;
}
}

int main(){
int n, m, u, v, l;
char ch;
while(cin >> n >> m){
init(n);
while(m--){
scanf("%d %d %d %c", &u, &v, &l, &ch);
G[u].push_back(v);  w[u].push_back(l);
G[v].push_back(u);  w[v].push_back(l);
}

int ans = 0;
for(int i = 1; i <= n; ++i)
if(!vvis[i])  ans = max(ans, d[bfs(bfs(i))]);//两次BFS，第一次是找最远的点，第二次是找最远点的最远点
cout << ans << endl;
}
return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>

using namespace std;
const int maxn = 1e5 + 5;
int f[maxn], g[maxn], ll[maxn];
vector<int> G[maxn], w[maxn];
//思路主要是找一个点的最远点加次远点就是树的直径

int dfs(int root, int fa){
if(f[root] != -1)  return f[root];
if(!G[root].size())  return f[root] = 0;
int m = 0, ans = root;
for(int i = 0; i < G[root].size(); ++i){
int u = G[root][i];
if(u == fa)  continue;
if(dfs(u, root) + w[root][i] > m){
m = f[u] + w[root][i];
ans = u;
}
}

ll[root] = ans; int  mm = 0;
for(int i = 0; i < G[root].size(); ++i){
int u = G[root][i];
if(u == fa)  continue;
if(f[u] + w[root][i] > mm && u != ll[root])
mm = f[u] + w[root][i];
}
g[root] = mm;
return f[root] = m;
}

void init(int n){
for(int i = 1; i <= n; ++i){
G[i].clear();
w[i].clear();
f[i] = g[i] = ll[i] = -1;
}
}

int main(){
int n, m, u, v, l;
char ch;
while(cin >> n >> m){
init(n);
while(m--){
scanf("%d %d %d %c", &u, &v, &l, &ch);
G[u].push_back(v);  w[u].push_back(l);
G[v].push_back(u);  w[v].push_back(l);
}

int ans = 0;
for(int i = 1; i <= n; ++i)
if(f[i] == -1)  dfs(i, -1);
for(int i = 1; i <= n; ++i)  ans = max(ans, f[i]+g[i]);
cout << ans << endl;
}
return 0;
}