# 문제 분류
최대 유량
최대 유량을 사용하면 되는 문제이다. 그리고 같은 간선에 유량이 증가하는 입력이 존재할 수 있으므로, 유량을 더하는 방식으로 입력을 받아야 한다.
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <queue>
#include <utility>
#include <vector>
#define mp make_pair
#define pb push_back
#define X first
#define Y second
#define fup(i, a, b, c) for (int(i) = (a); (i) <= (b); (i) += (c))
#define fdn(i, a, b, c) for (int(i) = (a); (i) >= (b); (i) -= (c))
#define endl "\n"
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pl;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const ll MOD = 1e9 + 7;
const int stMAX = 1 << 18;
const int INF = 1e9;
const int MAX_V = 60;
int c[MAX_V][MAX_V], f[MAX_V][MAX_V];
int level[MAX_V];
int work[MAX_V];
vi adj[MAX_V];
int func(char cap) {
if (cap <= 'Z')
return cap - 'A';
return cap - 'a' + 26;
}
int S = func('A'), E = func('Z');
bool bfs() {
fill(level, level + MAX_V, -1);
level[S] = 0;
queue<int> Q;
Q.push(S);
while (!Q.empty()) {
int curr = Q.front();
Q.pop();
for (int next : adj[curr]) {
if (level[next] == -1 && c[curr][next] - f[curr][next] > 0) {
level[next] = level[curr] + 1;
Q.push(next);
}
}
}
return level[E] != -1;
}
int dfs(int curr, int dest, int flow) {
if (curr == dest)
return flow;
for (int &i = work[curr]; i < adj[curr].size(); i++) {
int next = adj[curr][i];
if (level[next] == level[curr] + 1 && c[curr][next] - f[curr][next] > 0) {
int df = dfs(next, dest, min(c[curr][next] - f[curr][next], flow));
if (df > 0) {
f[curr][next] += df;
f[next][curr] -= df;
return df;
}
}
}
return 0;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int N;
cin >> N;
fup(_, 1, N, 1) {
char u, v;
int w;
cin >> u >> v >> w;
u = func(u);
v = func(v);
adj[u].pb(v);
adj[v].pb(u);
c[u][v] = c[v][u] += w;
}
int total = 0;
while (bfs()) {
fill(work, work + MAX_V, 0);
while (1) {
int flow = dfs(S, E, INF);
if (flow == 0)
break;
total += flow;
}
}
cout << total;
}