BZOJ 3363 POJ 1985 Cow Marathon 樹的直徑
阿新 • • 發佈:2017-07-21
ng- edge using -a scanf turn stream zoj article
題目大意:給出一棵樹。求兩點間的最長距離。
思路:裸地樹的直徑。兩次BFS,第一次隨便找一個點寬搜。然後用上次寬搜時最遠的點在寬搜。得到的最長距離就是樹的直徑。
CODE:
#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 80010 using namespace std; int points,edges; int head[MAX],total; int next[MAX],aim[MAX],length[MAX]; int f[MAX]; char s[10]; inline void Add(int x,int y,int len); inline void BFS(int start); int main() { cin >> points >> edges; for(int x,y,len,i = 1;i <= edges; ++i) { scanf("%d%d%d%s",&x,&y,&len,s); Add(x,y,len),Add(y,x,len); } BFS(1); int _max = 0; for(int i = 1;i <= points; ++i) if(f[i] > f[_max]) _max = i; BFS(_max); int ans = 0; for(int i = 1;i <= points; ++i) if(f[i] > ans) ans = f[i]; cout << ans << endl; return 0; } inline void Add(int x,int y,int len) { next[++total] = head[x]; aim[total] = y; length[total] = len; head[x] = total; } inline void BFS(int start) { static queue<int> q; while(!q.empty()) q.pop(); memset(f,0x3f,sizeof(f)); f[0] = f[start] = 0; q.push(start); while(!q.empty()) { int x = q.front(); q.pop(); for(int i = head[x];i;i = next[i]) if(f[aim[i]] > f[x] + length[i]) { f[aim[i]] = f[x] + length[i]; q.push(aim[i]); } } }
BZOJ 3363 POJ 1985 Cow Marathon 樹的直徑