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HDU 4514 湫湫系列故事――设计风景线 (树形DP)

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;}

 

HDU 4514 湫湫系列故事――设计风景线 (树形DP)