#include<stdio.h> #include<string.h> #define N 300 #define inf 0x3fffffff int mark[N],link[N],n,m; int map[N][N]; int find(int u) { int i; for(i=1;i<=m;i++) {     if(!mark[i]&&map[u][i]) {         mark[i]=1;         if(link[i]==-1||…

The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24081   Accepted: 10695 Description Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering p…

Luogu 1894 [USACO4.2]完美的牛栏The Perfect Stall / POJ 1274 The Perfect Stall(二分图最大匹配) Description 农夫约翰上个星期刚刚建好了他的新牛棚,他使用了最新的挤奶技术.不幸的是,由于工程问题,每个牛栏都不一样.第一个星期,农夫约翰随便地让奶牛们进入牛栏,但是问题很快地显露出来:每头奶牛都只愿意在她们喜欢的那些牛栏中产奶.上个星期,农夫约翰刚刚收集到了奶牛们的爱好的信息(每头奶牛喜欢在哪些牛栏产奶).一个牛栏只能容纳…

题目链接: http://poj.org/problem?id=3041 Description Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently…

过山车 Time Limit: 1000 MS Memory Limit: 32768 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main Description RPG girls今天和大家一起去游乐场玩,终于可以坐上梦寐以求的过山车了.可是,过山车的每一排只有两个座位,而且还有条不成文的规矩,就是每个女生必须找个个男生做partner和她同坐.但是,每个女孩都有各自的想法,举个例子把,Rabbit只愿意和XHD或P…

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13262   Accepted: 6412 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…

poj 2049(二分+spfa判负环) 给你一堆字符串,若字符串x的后两个字符和y的前两个字符相连,那么x可向y连边.问字符串环的平均最小值是多少.1 ≤ n ≤ 100000,有多组数据. 首先根据套路,二分是显然的.然后跑一下spfa判断正环就行了. 然而我被no solution坑了十次提交.. #include <cctype> #include <cstdio> #include <cstring> using namespace std; const in…

poj——1274   The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 25709   Accepted: 11429 Description Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to e…

/*1A 31ms*/ #include<stdio.h> #include<string.h> #define N 300 int n; struct node { int u,v,next; }bian[N*N*2]; int color[N],vis[N],link[N],visit[N],ma[N][N],f[N],head[N],yong; void addedge(int u,int v) { bian[yong].u=u; bian[yong].v=v; bian[y…

两题二分图匹配的题: 1.一个农民有n头牛和m个畜栏,对于每个畜栏,每头牛有不同喜好,有的想去,有的不想,对于给定的喜好表,你需要求出最大可以满足多少头牛的需求. 2.给你学生数和课程数,以及学生上的课,如果可以做到每个学生代表不同的课程并且所有的课程都被代表输出"YES"(学生能代表一门课当且仅当他上过). 1.POJ 1274 The Perfect Stall http://poj.org/problem?id=1274 和上一题过山车一样,也是二分图匹配的. 水题. #incl…