题目大意:给定对手的一副牌,现在你可以重新安排你的手牌顺序,最大化你的牌比对手的牌大的数量 ,类似于田忌赛马问题
题解:这道题还是一道二分匹配的问题,当你的牌比对手的牌大时,在这两张牌之间连接一条边,求最大匹配。
#include<bits/stdc++.h>
using namespace std;
int n;
const int maxn = 1500;
int scoreadam[maxn], scoreeve[maxn];
int edge[maxn][maxn], vis[maxn], link[maxn];
int getScore(string str) {//将牌转换为分数
int ans = 0;
if (str[0] >= '2' && str[0] <= '9')
ans += (str[0] - '0') * 10;
else if (str[0] == 'T')
ans += 100;
else if (str[0] == 'J')
ans += 110;
else if (str[0] == 'Q')
ans += 120;
else if (str[0] == 'K')
ans += 130;
else if (str[0] == 'A')
ans += 140;
if (str[1] == 'C')
ans += 1;
else if (str[1] == 'D')
ans += 2;
else if (str[1] == 'S')
ans += 3;
else if (str[1] == 'H')
ans += 4;
return ans;
}
int dfs(int l)
{
for (int r = 1; r <= n; r++)
{
if (edge[l][r]==1 && !vis[r])
{
vis[r] = 1;
if (link[r] == -1 || dfs(link[r]))
{
link[r] = l;
return 1;
}
}
}
return 0;
}
int hungary()
{
int cnt = 0;
memset(link, -1, sizeof(link));
for (int i = 1; i <= n; i++)
{
memset(vis, 0, sizeof(vis));
if (dfs(i))cnt++;
}
return cnt;
}
int main()
{
int t;
string s;
cin >> t;
while (t--)
{
memset(edge, 0, sizeof(edge));
memset(scoreadam, 0, sizeof(scoreadam));
memset(scoreeve, 0, sizeof scoreeve);
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> s;
scoreadam[i] = getScore(s);
}
for (int i = 1; i <= n; i++)
{
cin >> s;
scoreeve[i] = getScore(s);
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (scoreeve[i] > scoreadam[j])
edge[i][j] = 1;
}
}
cout << hungary() << endl;
}
}