http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1986
| My T-shirt suits me |
Our friend Victor participates as an instructor in an environmental volunteer program. His boss asked Victor to distribute N T-shirts
to Mvolunteers, one T-shirt each volunteer, where N is multiple of six, and N
M.
There are the same number of T-shirts of each one of the six available sizes: XXL, XL, L, M , S, and XS. Victor has a little problem because only two sizes of the T-shirts suit each volunteer.
You must write a program to decide if Victor can distribute T-shirts in such a way that all volunteers get a T-shirt that suit them. If N 
M,
there can be some remaining T-shirts.
The first line of the input contains the number of test cases. For each test case, there is a line with two numbers N and M. N is
multiple of 6, 1
N36,
and indicates the number of T-shirts. Number M, 1M30,
indicates the number of volunteers, with NM.
Subsequently, M lines are listed where each line contains, separated by one space, the two sizes that suit each volunteer (XXL, XL, L, M , S, or XS).
For each test case you are to print a line containing YES if there is, at least, one distribution where T-shirts suit all volunteers, or NO, in other case.
3 18 6 L XL XL L XXL XL S XS M S M L 6 4 S XL L S L XL L XL 6 1 L M
YES NO YES
题意：
有n(n是6的倍数)件衣服，6种尺码，每种尺码的衣服数量相同，有m个人，每人有两种能穿的尺码，问每个人是否都有衣服穿。
分析：
显然的二分图。每个人向其合适的尺码连边，容量为1;增加源点和汇点，源点向每个人连边，容量为1,每种尺码向汇点连边，容量为该种尺码衣服的数量(n/6)。在上图中跑最大流，如果满流则所有人都有衣服穿。
/*
*
* Author : fcbruce
*
* Date : 2014-09-04 20:47:21
*
*/
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#ifdef _WIN32
#define lld "%I64d"
#else
#define lld "%lld"
#endif
#define maxm 233
#define maxn 64
using namespace std;
int fir[maxn];
int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm];
int e_max;
int iter[maxn],q[maxn],lv[maxn];
void add_edge(int _u,int _v,int _w)
{
int e;
e=e_max++;
u[e]=_u;v[e]=_v;cap[e]=_w;
nex[e]=fir[u[e]];fir[u[e]]=e;
e=e_max++;
u[e]=_v;v[e]=_u;cap[e]=0;
nex[e]=fir[u[e]];fir[u[e]]=e;
}
void dinic_bfs(int s)
{
int f,r;
memset(lv,-1,sizeof lv);
q[f=r=0]=s;
lv[s]=0;
while(f<=r)
{
int x=q[f++];
for (int e=fir[x];~e;e=nex[e])
{
if (cap[e]>flow[e] && lv[v[e]]<0)
{
lv[v[e]]=lv[u[e]]+1;
q[++r]=v[e];
}
}
}
}
int dinic_dfs(int _u,int t,int _f)
{
if (_u==t) return _f;
for (int &e=iter[_u];~e;e=nex[e])
{
if (cap[e]>flow[e] && lv[_u]<lv[v[e]])
{
int _d=dinic_dfs(v[e],t,min(_f,cap[e]-flow[e]));
if (_d>0)
{
flow[e]+=_d;
flow[e^1]-=_d;
return _d;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
memset(flow,0,sizeof flow);
int total_flow=0;
for (;;)
{
dinic_bfs(s);
if (lv[t]<0) return total_flow;
memcpy(iter,fir,sizeof iter);
int _f;
while ((_f=dinic_dfs(s,t,INF))>0)
total_flow+=_f;
}
return total_flow;
}
char _size[7][5]={ "","XS","S","M","L","XL","XXL"};
int main()
{
#ifdef FCBRUCE
freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE
int T_T;
int n,m;
scanf( "%d",&T_T);
while (T_T--)
{
e_max=0;
memset(fir,-1,sizeof fir);
scanf( "%d%d",&n,&m);
int s=0,t=m+7;
char s1[5],s2[5];
for (int i=0;i<m;i++)
{
add_edge(s,i+7,1);
scanf( "%s%s",s1,s2);
for (int j=1;j<7;j++)
if (strcmp(s1,_size[j])==0) add_edge(i+7,j,1);
for (int j=1;j<7;j++)
if (strcmp(s2,_size[j])==0) add_edge(i+7,j,1);
}
for (int i=1;i<7;i++)
add_edge(i,t,n/6);
if (max_flow(s,t)==m)
puts( "YES");
else
puts( "NO");
}
return 0;
}
UVA 11045 My T-shirt suits me (二分图)
原文地址：http://blog.csdn.net/u012965890/article/details/39059321
