一次舞会有n个男孩和n个女孩。每首曲子开始时,所有男孩和女孩恰好配成n对跳交谊舞。每个男孩都不会和同一个女孩跳两首(或更多)舞曲。有一些男孩女孩相互喜欢,而其他相互不喜欢(不会“单向喜欢”)。每个男孩最多只愿意和k个不喜欢的女孩跳舞,而每个女孩也最多只愿意和k个不喜欢的男孩跳舞。给出每对男孩女孩是否相互喜欢的信息,舞会最多能有几首舞曲?
Input
第一行包含两个整数n和k。以下n行每行包含n个字符,其中第i行第j个字符为'Y'当且仅当男孩i和女孩j相互喜欢。
Output
仅一个数,即舞曲数目的最大值。
Sample Input
3 0 YYY YYY YYY
Sample Output
3
Hint
N<=50 K<=30
将每个人裂成 like , dislike ;
建立一个源点 s ,汇点 t;
由于求的是最大,不妨二分答案;
设此时答案为 m;
s 向每个男生的 like 连 m 的容量,每个男生的 like 与 dislike 连 k 的容量,
同样 ,每个女生的 dislike 与 like 连 k 容量,like 与 t 连m 容量;
此时对于一对喜欢的男女,他们的 like 连 1 容量,意味着只能跳一支舞;
如果相互不喜欢,其 dislike 连 1 容量;
dinic 跑一下最大流;
如果此时的最大流 Max_flow== n*m;
意味着满足,否则不满足;
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 999999999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
int n, k, tot;
int head[maxn << 2];
struct node {
int v, w, nxt;
}edge[maxn<<2];
int dac[200][200];
void addedge(int u, int v, int w) {
edge[++tot].v = v; edge[tot].w = w; edge[tot].nxt = head[u];
head[u] = tot;
}
int s, t;
int rk[maxn<<2];
int bfs() {
queue<int>q;
ms(rk);
rk[s] = 1; q.push(s);
while (!q.empty()) {
int tmp = q.front(); q.pop();
for (int i = head[tmp]; i; i = edge[i].nxt) {
int v = edge[i].v;
if (rk[v] || edge[i].w <= 0)continue;
rk[v] = rk[tmp] + 1; q.push(v);
}
}
return rk[t];
}
int dfs(int u, int flow) {
if (u == t)return flow;
int add = 0;
for (int i = head[u]; i&&add < flow; i = edge[i].nxt) {
int v = edge[i].v;
if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
int tmpadd = dfs(v, min(flow - add, edge[i].w));
if (tmpadd == 0) { rk[v] = -1; continue; }
edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd;
}
return add;
}
void construct(int a) {
ms(head); tot = 1;
for (int i = 1; i <= n; i++) {
addedge(s, i, a); addedge(i, s, 0);
addedge(i, i + n, k); addedge(i + n, i, 0);
addedge(i + 2 * n, i + 3 * n, k);
addedge(i + 3 * n, i + 2 * n, 0);
addedge(i + 3 * n, t, a); addedge(t, i + 3 * n, 0);
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (dac[i][j]) {
addedge(i, j + 3 * n, 1);
addedge(j + 3 * n, i, 0);
}
else {
addedge(i + n, j + 2 * n, 1);
addedge(j + 2 * n, i + n, 0);
}
}
}
}
int dinic(int mid) {
construct(mid);
int ans = 0;
while (bfs())ans += dfs(s, inf);
if (ans == mid * n)return true;
return false;
}
int log_search(int l, int r) {
int ans;
while (l <= r) {
int mid = (l + r) / 2;
if (dinic(mid))l = mid + 1, ans = mid;
else r = mid - 1;
}
return ans;
}
int main()
{
//ios::sync_with_stdio(false);
rdint(n); rdint(k);
char ch;
for(int i=1;i<=n;i++)
for (int j = 1; j <= n; j++) {
cin >> ch;
if (ch == 'Y')dac[i][j] = 1;
}
s = 4 * n + 1; t = s + 1;
cout << log_search(0, n + k) << endl;
return 0;
}