HDU 4514 湫湫系列故事――设计风景线 (树形DP)

Posted 2020-08-12 dwtfukgv

题意：略。

析：首先先判环，如果有环直接输出，用并查集就好，如果没有环，那么就是一棵树，然后最长的就是树的直径，这个题注意少开内存，容易超内存，

还有用C++交用的少一些，我用G++交的卡在32764K，限制是32768K。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<P> G[maxn];
int p[maxn], dp[maxn];
bool vis[maxn], viss[maxn];

int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]);  }

int bfs(int root){
    memset(vis, false, sizeof vis);
    memset(dp, 0, sizeof dp);
    queue<int> q;
    q.push(root);
    vis[root] = viss[root] = true;
    int ans = root, maxx = 0;

    while(!q.empty()){
        int u = q.front();  q.pop();
        for(int i = 0; i < G[u].size(); ++i){
            P p = G[u][i];
            int v = p.first;
            int w = p.second;
            if(vis[v])  continue;
            vis[v] = viss[v] = true;
            dp[v] = dp[u] + w;
            if(maxx < dp[v]){
                maxx = dp[v];
                ans = v;
            }
            q.push(v);
        }
    }
    return ans;
}

int solve(int root){
    int u = bfs(root);
    int v = bfs(u);
    return dp[v];
}

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        int u, v, c;
        for(int i = 1; i <= n; ++i)  G[i].clear(), p[i] = i;
        bool ok = false;
        for(int i = 0; i < m; ++i){
            scanf("%d %d %d", &u, &v, &c);
            int x = Find(u);
            int y = Find(v);
            if(x != y)  p[y] = x;
            else ok = true;
            G[u].push_back(P(v, c));
            G[v].push_back(P(u, c));
        }
        if(ok){  puts("YES");  continue; }

        memset(viss, false, sizeof viss);
        int ans = 0;
        for(int i = 1; i <= n; ++i)
            if(!viss[i])  ans = Max(ans , solve(i));
        printf("%d\n", ans);
    }
    return 0;
}