POJ2584

来源：互联网发布：工厂下料软件编辑：程序博客网时间：2024/05/22 03:29

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;}

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