2016-2017 ACM-ICPC, NEERC, Central Subregional Contest J.Architect of Your Own Fortune

题目大意:寻找luckynumber，她的定义是从n和m中各挑选一个六位数，使之n中的前三位之和与m中的后三位之和相同或者n中的后三位之和与m中的前三位之和相同，已经被选中的无法再组合，要是上面两种情况同时符合，选择和最大的一组，要是前者符合条件，在输出的时候输出AT和选中的两个六位数，否则是TA选中的两个六位数，排在前面的优先输出，然后按照第一个六位数的降序排列

解题思路:这是一道二分图的题，将符合条件的进行连线，然后按照匈牙利算法进行二分图匹配，之后遍历排序答案即可

#include<iostream>
#include<cstdio>
#include<stdio.h>
#include<cstring>
#include<cstdio>
#include<climits>
#include<cmath>
#include<vector>
#include <bitset>
#include<algorithm>
#include <queue>
#include<map>
#define inf 9999999;
using namespace std;

/*start:2017/5/10 13:15*/
/*end:2017/5/10 13:58*/

struct luckynumber
{
string str;
int x, y;
}ln[2000];
vector<int> tt[105];
bool vis[105];
int tried[105], at[105];
int a[105], b[105], n, m;
bool cmp(luckynumber x, luckynumber y)
{
return x.x > y.x;
}
bool check(int x, int y)
{
int la, ra, lb, rb;
la = (a[x] / 1000)%10+(a[x]/10000)%10+(a[x]/100000)%10;
ra = (a[x] % 1000)%10+(a[x]%1000)/10%10+(a[x]%1000)/100%10;
lb = (b[y] / 1000) % 10 + (b[y] / 10000) % 10 + (b[y] / 100000) % 10;
rb = (b[y] % 1000) % 10 + (b[y] % 1000) / 10 % 10 + (b[y] % 1000) / 100 % 10;
if (la == rb || ra == lb)
{
return true;
}
else
{
return false;
}
}
void flag(int x, int y)
{
int la, ra, lb, rb;
la = (a[x] / 1000) % 10 + (a[x] / 10000) % 10 + (a[x] / 100000) % 10;
ra = (a[x] % 1000) % 10 + (a[x] % 1000) / 10 % 10 + (a[x] % 1000) / 100 % 10;
lb = (b[y] / 1000) % 10 + (b[y] / 10000) % 10 + (b[y] / 100000) % 10;
rb = (b[y] % 1000) % 10 + (b[y] % 1000) / 10 % 10 + (b[y] % 1000) / 100 % 10;
if (la == rb&&ra == lb)
{
if (ra > la)
{
at[y] = 1;
}
else
{
at[y] = -1;
}
}
else if (la =
4000
= rb)
{
at[y] = -1;
}
else
{
at[y] = 1;
}
}
bool dfs(int x)
{
for (int i = 0; i < tt[x].size(); i++)
{
if (vis[tt[x][i]] == 0)
{
vis[tt[x][i]] = 1;
if (tried[tt[x][i]] == 0 || dfs(tried[tt[x][i]]))
{
flag(x, tt[x][i]);
tried[tt[x][i]] = x;
return true;
}
}
}
return false;
}
int main()
{
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
cin >> n >> m;
int i, j, ans;
for (i = 1; i <= n; i++)
{
cin >> a[i];
}
for (i = 1; i <= m; i++)
{
cin >> b[i];
}
for (i = 1; i <= n; i++)
{
for (j = 1; j <= m; j++)
{
if (check(i, j))
{
tt[i].push_back(j);
}
}
}
memset(vis, 0, sizeof(vis));
memset(tried, 0, sizeof(tried));
memset(at, 0, sizeof(at));
for (i = 1; i <= n; i++)
{
memset(vis, 0, sizeof(vis));
dfs(i);
}
ans = 0;
for (i = 1; i <= m; i++)
{
if (tried[i] != 0)
{
ans++;
if (at[i] == 1)
{
ln[ans].str = "TA";
ln[ans].x = b[i];
ln[ans].y =a[tried[i]] ;
}
else
{
ln[ans].str = "AT";
ln[ans].x = a[tried[i]];
ln[ans].y = b[i];
}
}
}
sort(ln+1, ln+1 + ans, cmp);
cout << ans << endl;
for (i = 1; i <= ans; i++)
{
cout << ln[i].str << " " << ln[i].x << " " << ln[i].y << endl;
}
}