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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 NM. 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.

Input

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, 1N36, 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).

Output

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.

Sample Input

 
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

Sample Output

 
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 
 *
 */
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                   #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]
                  
                   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
                   
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