题目大意：太暴力了，就不写了，看这儿
题解：对于每个$byx$的人，从源点向人连边，容量为此人的寿命。
对于每个手气君的人，从人向汇点连边，容量为此人的寿命。
对于每个$byx$的人与手气君的人，如果$byx$能够用此人赢手气君，从$byx$的这个人向手气君的这个人连一条边，容量为$1$。
对于长者，他的生命要加上本方膜法师的人数，代表续命。
卡点：1.忘记开反向弧（这样也有$10$分，果然暴力）
2.数组开小
C++ Code：
#include <cstdio>
#include <cstring>
#define maxn 111 << 1
#define maxm 111 * 111
const int inf = 0x3f3f3f3f;
int S[5][5] = {
{0, 1, 1, 0, 0},
{0, 0, 1, 1, 0},
{0, 0, 0, 1, 1},
{1, 0, 0, 0, 1},
{1, 1, 0, 0, 0}
};
int st, ed;
inline int min(int a, int b) {return a < b ? a : b;}
namespace Dinic {
int st, ed, sz;
int head[maxn], cnt = 2;
struct Edge {
int to, nxt, w;
} e[maxm << 1];
inline void add(int a, int b, int c) {
e[cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
e[cnt ^ 1] = (Edge) {a, head[b], 0}; head[b] = cnt ^ 1;
cnt += 2;
}
int d[maxn], q[maxn], h, t;
inline bool bfs() {
memset(d, 0, sz);
d[q[h = t = 0] = st] = 1;
while (h <= t) {
int u = q[h++];
// printf("u:%d\n", u);
if (u == ed) return true;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (!d[v] && e[i].w) {
d[v] = d[u] + 1;
// printf("v: %d %d\n", v, d[v]);
q[++t] = v;
}
}
}
// printf("d[ed]:%d\n", d[ed]);
return d[ed];
}
int dfs(int x, int low) {
if (!low || x == ed) return low;
int res = 0, w;
for (int i = head[x]; i; i = e[i].nxt) {
int v = e[i].to;
if (d[v] == d[x] + 1 && e[i].w) {
w = dfs(v, min(e[i].w, low - res));
res += w;
e[i].w -= w;
e[i ^ 1].w += w;
if (res == low) return low;
}
}
if (!res) d[x] = -1;
return res;
}
inline int dinic(int ST, int ED) {
st = ST, ed = ED, sz = sizeof d;
int ans = 0;
while (bfs()) ans += dfs(st, inf); //, printf("%d\n", ans);
return ans;
}
}
int n, m, A, B;
inline int get(char *s) {
if (*s == 'J') return 0;
if (*s == 'H') return 1;
if (*s == 'W') return 2;
if (*s == 'E') return 3;
if (*s == 'Y') return 4;
return 20040826;
}
char a[maxn][10], b[maxn][10];
int main() {
scanf("%d%d", &n, &m);
st = 0, ed = n << 1 | 1;
for (int i = 1; i <= n; i++) scanf("%s", a[i]), A += *a[i] == 'Y';
for (int i = 1; i <= n; i++) scanf("%s", b[i]), B += *b[i] == 'Y';
for (int i = 1, x; i <= n; i++) {
scanf("%d", &x);
Dinic::add(st, i, x + (*a[i] == 'J' ? A : 0));
}
for (int i = 1, x; i <= n; i++) {
scanf("%d", &x);
Dinic::add(i + n, ed, x + (*b[i] == 'J' ? B : 0));
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (S[get(a[i])][get(b[j])]) Dinic::add(i, j + n, 1); //, printf("%d %d :%c %c\n", i, j, *a[i], *b[j]);
}
}
printf("%d\n", min(Dinic::dinic(st, ed), m));
return 0;
}
