[题目链接] 点击打开链接 [算法] 二分答案,check的时候跑最大流,即可 [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 2000 #define MAXB 1000 int i,j,low,high,mid,st,ed,N,B,tot,ans,tmp; int h[MAXN+MAXB],U[MAXN*MAXB],V[MAXN*MAXB], W[MAXN*MAXB],Head[MAXN*MAXB],Next[M…

 Steady Cow Assignment Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3189 Description Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of co…

Steady Cow Assignment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6817   Accepted: 2349 Description Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. So…

题目链接:https://vjudge.net/problem/POJ-3189 Steady Cow Assignment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6979   Accepted: 2418 Description Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns wh…

Steady Cow Assignment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7482   Accepted: 2572 题目链接:http://poj.org/problem?id=3189 Description: Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which,…

题目链接: Poj 3189 Steady Cow Assignment 题目描述: 有n头奶牛,m个棚,每个奶牛对每个棚都有一个喜爱程度.当然啦,棚子也是有脾气的,并不是奶牛想住进来就住进来,超出棚子的最大容量了,棚子是拒绝的.现在要给每个奶牛安家,本宝宝是一个公正无私的人类,所以要找一个奶牛喜爱程度差值最小的方案(虽然这样大家的喜爱程度可能普遍偏低,因为绝对公平并不代表合理啊),问喜爱程度的区间最小为多大? 解题思路: 每个棚子并不是住一个奶牛,所以是多重匹配咯.匹配的时候二分枚举喜爱程度的…

1434:[例题2]Best Cow Fences 时间限制: 1000 ms         内存限制: 65536 KB提交数: 263     通过数: 146 [题目描述] 给定一个长度为n的正整数序列A.求一个平均数最大的,长度不小于L的子序列. [输入] 第一行,n和L: n个正整数,表示A. [输出] 一个整数,表示答案的1000倍(不用四舍五入,直接输出). [输入样例] 10 6 6 4 2 10 3 8 5 9 4 1 [输出样例] 6500 [提示] n ≤ 10000 思…

[题目链接] 点击打开链接 [算法] 树状数组 [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 100010 int n,i,x; long long ans; int pos[MAXN]; long long c[MAXN]; inline int lowbit(int x) { return x & (-x); } inline void modify(int pos,long long val) { i…

[题目链接] 点击打开链接 [算法] dist[i][j][k]表示当前走到(i,j),走的步数除以3的余数为k的最小花费 spfa即可 [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 110 const int INF = 1e9; struct info { int x,y,s; }; ] = {,,-,}; ] = {-,,,}; int i,j,n,t; int val[MAXN][MAXN]; templa…

[题目链接] 点击打开链接 [算法] SPFA + SLF / LLL 优化 [代码] #include<bits/stdc++.h> using namespace std; #define MAXT 25000 int i,T,R,P,S,u,v,w; ],vis[MAXT+]; vector<pair<]; template <typename T> void read(T &x) { ; ; ; } +c-'; x*=f; } inline void S…