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;

}