裸最短路.. ------------------------------------------------------------------------------------ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<queue>   #define rep( i , n ) for( int i = 0 ; i &…

因为是双向边,所以相当于两条到1的最短路和,先跑spfa然后直接处理询问即可 #include<iostream> #include<cstdio> #include<queue> using namespace std; const int N=50005,inf=1e9; int n,m,b,h[N],cnt,dis[N]; bool v[N]; struct qwe { int ne,to,va; }e[N<<2]; int read() { int…

2015: [Usaco2010 Feb]Chocolate Giving Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 269  Solved: 183[Submit][Status] Description Farmer John有B头奶牛(1<=B<=25000),有N(2*B<=N<=50000)个农场,编号1-N,有M(N-1<=M<=100000)条双向边,第i条边连接农场R_i和S_i(1<=R_i&…

http://www.lydsy.com/JudgeOnline/problem.php?id=2015 这种水题真没啥好说的.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace s…

Description Farmer John有B头奶牛(1<=B<=25000),有N(2*B<=N<=50000)个农场,编号1-N,有M(N-1<=M<=100000)条双向边,第i条边连接农场R_i和S_i(1<=R_i<=N;1<=S_i<=N),该边的长度是L_i(1<=L_i<=2000).居住在农场P_i的奶牛A(1<=P_i<=N),它想送一份新年礼物给居住在农场Q_i(1<=Q_i<=N)的…

[bzoj 1782] [Usaco2010 Feb]slowdown慢慢游 Description 每天Farmer John的N头奶牛(1 <= N <= 100000,编号1-N)从粮仓走向他的自己的牧场.牧场构成了一棵树,粮仓在1号牧场.恰好有N-1条道路直接连接着牧场,使得牧场之间都恰好有一条路径相连.第i条路连接着A_i,B_i,(1 <= A_i <= N; 1 <= B_i <= N).奶牛们每人有一个私人牧场P_i (1 <= P_i <=…

orz...hzwer 对着大神的 code 看 , 稍微理解了. 考虑一只牛到达 , 那它所在子树全部 +1 , 可以用BIT维护 ----------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<vector…

http://www.lydsy.com/JudgeOnline/problem.php?id=2014 这应该是显然的贪心吧,先排序,然后按花费取 #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using na…

跑两遍最短路就好了.. 话说这翻译2333 ---------------------------------------------------------------------- #include<cstdio> #include<queue> #include<algorithm> #include<cstring> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i…

这道题和蔡大神出的今年STOI初中组的第二题几乎一模一样... 先跑一遍最短路 , 再把所有边反向 , 再跑一遍 , 所有点两次相加的最大值即为answer ----------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<queue> #in…