1687: [Usaco2005 Open]Navigating the City 城市交通 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 122  Solved: 85[Submit][Status][Discuss] Description     由于牛奶市场的需求,奶牛必须前往城市,但是唯一可用的交通工具是出租车.教会奶牛如何在城市里打的.     给出一个城市地图,东西街区E(1≤E≤40),南北街区N(1≤N≤30).制作一个开车指南给出…
1687: [Usaco2005 Open]Navigating the City 城市交通 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 94  Solved: 73[Submit][Status][Discuss] Description     由于牛奶市场的需求,奶牛必须前往城市,但是唯一可用的交通工具是出租车.教会奶牛如何在城市里打的.     给出一个城市地图,东西街区E(1≤E≤40),南北街区N(1≤N≤30).制作一个开车指南给出租…
http://www.lydsy.com/JudgeOnline/problem.php?id=1687 bfs后然后逆向找图即可.因为题目保证最短路唯一 #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using…
Description A dip in the milk market has forced the cows to move to the city. The only employment available is in the venerable field of taxi-driving. Help the cows learn their way around the city. Given a city map with E (1 <= E <= 40) east/west st…
1674: [Usaco2005]Part Acquisition Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 337  Solved: 162[Submit][Status][Discuss] Description The cows have been sent on a mission through space to acquire a new milking machine for their barn. They are flying…
题目链接:BZOJ - 1733 题目分析 直接二分这个最大边的边权,然后用最大流判断是否可以有 T 的流量. 代码 #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MaxN = 200 + 5,…
题目 1739: [Usaco2005 mar]Space Elevator 太空电梯 Time Limit: 5 Sec  Memory Limit: 64 MB Description The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) differe…
题目 1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚 Time Limit: 5 Sec  Memory Limit: 64 MB Description Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most…
完全背包.. --------------------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; i++ ) #defi…
一开始直接 O( n² ) 暴力..结果就 A 了... USACO 数据是有多弱 = = 先sort , 然后自己再YY一下就能想出来...具体看code ----------------------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<i…