POJ2584

Problem: T-Shirt Gumbo
Description: 一群运动员他们对自己T恤的尺码有一个范围的要求。现在给出5种尺码的衣服的数量，问每个运动员是否都能选到自己满意的尺码的衣服。
Solution: 这个题一看就是二分图的多重匹配，但是我一开始是不知道的，于是我就用了二分图的最大匹配来做，我们不要看种类，直接看每一件衣服，这样就转化成了每个运动员都能选到衣服。
Code(最大匹配):

#include <stdio.h>
#include <string.h>

#include <iostream>
#include <string>
#include <vector>

using namespace std;

const int M = 200;

int n, m;
int num[10];

vector<int> map[25];
int belong[M];
bool used[M];

int getnum(char s)
{
    if (s == 'S')
        return 1;
    if (s == 'M')
        return 2;
    if (s == 'L')
        return 3;
    if (s == 'X')
        return 4;
    if (s == 'T')
        return 5;
    return 0;
}

bool dfs(int s)
{
    for (int i = 0; i < map[s].size(); i++) {
        int end = map[s].at(i);
        if (used[end])
            continue;
        used[end] = true;
        if (belong[end] == -1 || dfs(belong[end])) {
            belong[end] = s;
            return true;
        }
    }
    return false;
}

int main()
{
    string str;
    while (cin >> str, str.at(0) != 'E') {
        for (int i = 0; i < 25; i++)
            map[i].clear();
        cin>>n;
        string edge[25];
        int sum = 0, x;
        for (int i = 1; i <= n; i++)
            cin >> edge[i];
        for (int i = 1; i <= 5; i++)
            cin >> x,
            num[i] = sum, sum += x;
        num[6] = sum;
        cin>>str;
        if (sum < n) {
            cout << "I'd rather not wear a shirt anyway..." << endl;
            continue;
        }

        for (int i = 1; i <= n; i++) {
            int start = num[getnum(edge[i].at(0))];
            int end = num[getnum(edge[i].at(1)) + 1];
            for (int j = start + 1; j <= end; j++)
                map[i].push_back(j);
        }
        int ans = 0;
        memset(belong, -1, sizeof(belong));
        for (int i = 1; i <= n; i++) {
            memset(used, false, sizeof(used));
            if (dfs(i))
                ++ans;
        }
        if (ans == n)
            cout << "T-shirts rock!";
        else
            cout << "I'd rather not wear a shirt anyway...";
        cout << endl;
    }
    return 0;
}

Code(多重匹配):

#include <stdio.h>
#include <string.h>

#include <iostream>
#include <string>
#include <vector>

using namespace std;

const int M = 20;

int n, m;
int num[10];

vector<int> map[25];
int belong[M][M];
int used[M];

int getnum(char s)
{
    if (s == 'S')
        return 1;
    if (s == 'M')
        return 2;
    if (s == 'L')
        return 3;
    if (s == 'X')
        return 4;
    if (s == 'T')
        return 5;
    return 0;
}

bool dfs(int s)
{
    for (int i = 0; i < map[s].size(); i++) {
        int end = map[s].at(i);
        if (used[end] >= num[end])
            continue;
        ++used[end];
        for (int j = 1; j <= num[end]; j++)
            if (belong[end][j] == -1 || dfs(belong[end][j])) {
                belong[end][j] = s;
                return true;
            }
    }
    return false;
}

int main()
{
    string str;
    while (cin >> str, str.at(0) != 'E') {
        for (int i = 0; i < 25; i++)
            map[i].clear();
        cin>>n;
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            cin >> str;
            int start = getnum(str.at(0));
            int end = getnum(str.at(1));
            for (int j = start; j <= end; j++)
                map[i].push_back(j);
        }
        for (int i = 1; i <= 5; i++)
            cin >> num[i], sum += num[i];
        cin>>str;
        if (sum < n) {
            cout << "I'd rather not wear a shirt anyway..." << endl;
            continue;
        }
        int ans = 0;
        memset(belong, -1, sizeof(belong));
        for (int i = 1; i <= n; i++) {
            memset(used, 0, sizeof(used));
            if (dfs(i))
                ++ans;
        }
        if (ans == n)
            cout << "T-shirts rock!";
        else
            cout << "I'd rather not wear a shirt anyway...";
        cout << endl;
    }
    return 0;
}