POJ 2584 TShirt Gumbo(最大流)

题意：有n个同学要求订购衣服，每个同学可以接收的衣服型号已知（一个人一件衣服），现在每个型号的衣服数量已知，问能不能满足同学的要求？

思路：可行性问题，此题方法很多，多重二分匹配、BFS等，我偷懒用了最大流，BS我吧.......还是说下建图过程吧，增加源汇点s,t，每个同学与其所接受的型号连一条1的边，s与同学连INF边，型号和t连val (val 是此型号的数量)，然后求最大流......

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 200;
const int INF = INT_MAX;

#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

int index,nv,n,m,k,s,t;
int ans;
int net[MAXN];
int need[30];
int have[10];

struct Edge{
    int next,pair;
    int v,cap,flow;
}edge[MAXN*MAXN];

int init()
{
    CLR(net,-1);
    CLR(have,0);
    CLR(need,0);
    index = 0;
    need['S'] = 1;
    need['M'] = 2;
    need['L'] = 3;
    need['X'] = 4;
    need['T'] = 5;
    return 0;
}
void add_edge(const int& u,const int& v,const int& val)
{
    edge[index].next = net[u];
    net[u] = index;
    edge[index].v = v;
    edge[index].cap = val;
    edge[index].flow = 0;
    edge[index].pair = index+1;
    ++index;
    edge[index].next = net[v];
    net[v] = index; 
    edge[index].v = u;
    edge[index].cap = 0;
    edge[index].flow = 0;
    edge[index].pair = index - 1;
    ++index;
}
int input()
{
    string op;
    int i,j,start,end;
    cin>>op;
    if(op.length() > 6)
        return 1;
    cin>>n;
    s = n+5+1;
    t = n+5+2;
    nv = t;
    for(i = 1; i <= n; ++i)
    {
        cin>>op;
        for(j = need[op[0]]; j <= need[op[1]]; ++j)
            add_edge(i,n+j,1);
        add_edge(s,i,1);
    }
    for(i = 1; i <= 5; ++i)
    {
        cin>>j;
        add_edge(i+n,t,j);
    }
    cin>>op;
    return 0;
}
int ISAP()
{
    long numb[MAXN],dist[MAXN],curedge[MAXN],pre[MAXN];
    long cur_flow,max_flow,u,tmp,neck,i;
    memset(dist,0,sizeof(dist));
    memset(numb,0,sizeof(numb));
    for(i = 1 ; i <= nv ; ++i)
        curedge[i] = net[i];
    numb[nv] = nv;
    max_flow = 0;
    u = s;
    while(dist[s] < nv)
    {
        /* first , check if has augmemt flow */
        if(u == t)
        {
            cur_flow = INF;
            for(i = s; i != t;i = edge[curedge[i]].v) 
            {  
                if(cur_flow > edge[curedge[i]].cap)
                {
                    neck = i;
                    cur_flow = edge[curedge[i]].cap;
                }
            }
            for(i = s; i != t; i = edge[curedge[i]].v)
            {
                tmp = curedge[i];
                edge[tmp].cap -= cur_flow;
                edge[tmp].flow += cur_flow;
                tmp = edge[tmp].pair;
                edge[tmp].cap += cur_flow;
                edge[tmp].flow -= cur_flow;
            }

            max_flow += cur_flow;
            u = s;
        }
        /* if .... else ... */
        for(i = curedge[u]; i != -1; i = edge[i].next)
            if(edge[i].cap > 0 && dist[u] == dist[edge[i].v]+1)
                break;
        if(i != -1)
        {
            curedge[u] = i;
            pre[edge[i].v] = u;
            u = edge[i].v;
        }else{
            if(0 == --numb[dist[u]]) break;
            curedge[u] = net[u];
            for(tmp = nv,i = net[u]; i != -1; i = edge[i].next)
                if(edge[i].cap > 0)
                    tmp = tmp<dist[edge[i].v]?tmp:dist[edge[i].v];
            dist[u] = tmp + 1;
            ++numb[dist[u]];
            if(u != s) u = pre[u];
        }
    }
    return max_flow;
}
int work()
{
    int i,j,tmp;
    ans = ISAP();
    if(ans == n)
        printf("T-shirts rock!\n");
    else
        printf("I'd rather not wear a shirt anyway...\n");
    return 0;
}
int main()
{
    int res;
    while(true)
    {
        init();
        res = input();
        if(res)
            break;
        work();
    }
    return 0;
}