UVa 11045 - My T-shirt suits me（最大流）

题意

一个人有两种合适的T恤，现在有K套，问能不能让所有人穿上合适的。

思路

乍一看感觉是二分图匹配的内容？可是我不会。

可以这么想：建一个源点，到每个衣服的容量显然可以算出。每个衣服到每个合适的人的容量为1，每个人最后都连上汇点，这条路容量为1（因为只能穿一件）。如果最后流量能达到人数，说明可以，否则不可以。

把最大流封装了一下。

代码

  
  
  
  

   
   
   
   
#include <cstdio>
   
   
   
   
#include <stack>
   
   
   
   
#include <set>
   
   
   
   
#include <iostream>
   
   
   
   
#include <string>
   
   
   
   
#include <vector>
   
   
   
   
#include <queue>
   
   
   
   
#include <functional>
   
   
   
   
#include <cstring>
   
   
   
   
#include <algorithm>
   
   
   
   
#include <cctype>
   
   
   
   
#include <ctime>
   
   
   
   
#include <cstdlib>
   
   
   
   
#include <fstream>
   
   
   
   
#include <string>
   
   
   
   
#include <sstream>
   
   
   
   
#include <map>
   
   
   
   
#include <cmath>
   
   
   
   
#define LL long long
   
   
   
   
#define SZ(x) (int)x.size()
   
   
   
   
#define Lowbit(x) ((x) & (-x))
   
   
   
   
#define MP(a, b) make_pair(a, b)
   
   
   
   
#define MS(arr, num) memset(arr, num, sizeof(arr))
   
   
   
   
#define PB push_back
   
   
   
   
#define F first
   
   
   
   
#define S second
   
   
   
   
#define ROP freopen("input.txt", "r", stdin);
   
   
   
   
#define MID(a, b) (a + ((b - a) >> 1))
   
   
   
   
#define LC rt << 1, l, mid
   
   
   
   
#define RC rt << 1|1, mid + 1, r
   
   
   
   
#define LRT rt << 1
   
   
   
   
#define RRT rt << 1|1
   
   
   
   
#define BitCount(x) __builtin_popcount(x)
   
   
   
   
const double PI = acos(-1.0);
   
   
   
   
const int INF = 0x3f3f3f3f;
   
   
   
   
using namespace std;
   
   
   
   
const LL MAXN = 30 + 10;
   
   
   
   
const int MOD = 20071027;
   
   
   
   

   
   
   
   
typedef pair<int, int> pii;
   
   
   
   
typedef vector<int>::iterator viti;
   
   
   
   
typedef vector<pii>::iterator vitii;
   
   
   
   

   
   
   
   
int Convert(string str)
   
   
   
   
{
   
   
   
   
 if (str == "XXL") return 1;
   
   
   
   
 else if (str == "XL") return 2;
   
   
   
   
 else if (str == "L") return 3;
   
   
   
   
 else if (str == "M") return 4;
   
   
   
   
 else if (str == "S") return 5;
   
   
   
   
 else if (str == "XS") return 6;
   
   
   
   
}
   
   
   
   

   
   
   
   
​struct EDGE
   
   
   
   
{
   
   
   
   
 int from, to, cap, flow;
   
   
   
   
};
   
   
   
   

   
   
   
   
struct MAXFLOW
   
   
   
   
{
   
   
   
   
 int a[MAXN], p[MAXN], N;
   
   
   
   
 vector<EDGE> edge;
   
   
   
   
 vector<int> G[MAXN];
   
   
   
   

   
   
   
   
 void init(int n)
   
   
   
   
 {
   
   
   
   
 this->N = n;
   
   
   
   
 for (int i = 0; i <= n; i++) G[i].clear();
   
   
   
   
 }
   
   
   
   

   
   
   
   
 void add_edge(int from, int to, int cap)
   
   
   
   
 {
   
   
   
   
 edge.PB((EDGE){from, to, cap, 0});
   
   
   
   
 edge.PB((EDGE){to, from, 0, 0});
   
   
   
   
 int m = SZ(edge);
   
   
   
   
 G[from].PB(m - 2);
   
   
   
   
 G[to].PB(m - 1);
   
   
   
   
 }
   
   
   
   

   
   
   
   
 void EK(int st, int &flow)
   
   
   
   
 {
   
   
   
   
 queue<int> qu;
   
   
   
   
 while (true)
   
   
   
   
 {
   
   
   
   
 MS(a, 0);
   
   
   
   
 a[0] = INF;
   
   
   
   
 qu.push(0);
   
   
   
   
 while (!qu.empty())
   
   
   
   
 {
   
   
   
   
 int u = qu.front(); qu.pop();
   
   
   
   
 for (int i = 0; i < SZ(G[u]); i++)
   
   
   
   
 {
   
   
   
   
 EDGE &e = edge[G[u][i]];
   
   
   
   
 if (!a[e.to] && e.cap > e.flow)
   
   
   
   
 {
   
   
   
   
 qu.push(e.to);
   
   
   
   
 p[e.to] = G[u][i];
   
   
   
   
 a[e.to] = min(a[u], e.cap - e.flow);
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 if (!a[N]) break;
   
   
   
   
 int u = N;
   
   
   
   
 while (u != st)
   
   
   
   
 {
   
   
   
   
 edge[p[u]].flow += a[N];
   
   
   
   
 edge[p[u] ^ 1].flow -= a[N];
   
   
   
   
 u = edge[p[u]].from;
   
   
   
   
 }
   
   
   
   
 flow += a[N];
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
}maxFlow;
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 //ROP;
   
   
   
   
 ios::sync_with_stdio(0);
   
   
   
   

   
   
   
   
 int T, i, j, nmax, npeo;
   
   
   
   
 cin >> T;
   
   
   
   
 while (T--)
   
   
   
   
 {
   
   
   
   
 cin >> nmax >> npeo;
   
   
   
   
 maxFlow.init(npeo + 6 + 1);
   
   
   
   
 int f = nmax / 6;
   
   
   
   
 string s, ss;
   
   
   
   
 for (i = 1; i <= npeo; i++)
   
   
   
   
 {
   
   
   
   
 cin >> s;
   
   
   
   
 int tmp = Convert(s);
   
   
   
   
 maxFlow.add_edge(tmp, i + 6, 1);
   
   
   
   
 cin >> s;
   
   
   
   
 tmp = Convert(s);
   
   
   
   
 maxFlow.add_edge(tmp, i + 6, 1);
   
   
   
   
 }
   
   
   
   
 for (i = 1; i <= 6; i++) maxFlow.add_edge(0, i, f);
   
   
   
   
 for (i = 7; i <= npeo + 6; i++) maxFlow.add_edge(i, npeo + 6 + 1, 1);
   
   
   
   
 f = 0;
   
   
   
   
 maxFlow.EK(0, f);
   
   
   
   
 if (f == npeo) puts("YES");
   
   
   
   
 else puts("NO");
   
   
   
   
 }
   
   
   
   
 return 0;
   
   
   
   
}