学步园

两道的题型都是：给出现有的匹配，让你求匹配的权值最大时，改动的边数。

方法就是把边上标记，上标记的方法就是把权值扩大比N大的一个倍数，比如说N：50，那么我们就扩大55倍，原来已经存在的匹配边权值再加1，就这样，无论有原有匹的匹配边有多少条，一定要小于55，通过这样又标记出了没有改变的匹配边。求得ans/55时，这个值就是最大的匹配权值，ans%55就是没有改变的匹配边。

/*HDU 2853*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cctype>
using namespace std;

const int maxn = 60;
const int INF = 0x3f3f3f3f;

int n, m;

int W[maxn][maxn];
int Lx[maxn], Ly[maxn];

int Left[maxn];
bool S[maxn], T[maxn];

bool match(int i)
{
    S[i] = 1;
    for(int j = 1; j <= m; j++) if(Lx[i]+Ly[j] == W[i][j] && !T[j])
    {
        T[j] = 1;
        if(!Left[j] || match(Left[j]))
        {
            Left[j] = i;
            return 1;
        }
    }
    return 0;
}

void update()
{
    int a = INF;
    for(int i = 1; i <= n; i++) if(S[i])
    for(int j = 1; j <= m; j++) if(!T[j])
        a = min(a, Lx[i]+Ly[j]-W[i][j]);
        
    for(int i = 1; i <= n; i++)
    {
        if(S[i]) Lx[i] -= a;
    }
    for(int j = 1; j <= m; j++)
    {
        if(T[j]) Ly[j] += a;
    }
}

void KM()
{
    memset(Left, 0, sizeof(Left));
    memset(Lx, 0, sizeof(Lx));
    memset(Ly, 0, sizeof(Ly));
    
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
            Lx[i] = max(Lx[i], W[i][j]);
    }
    for(int i = 1; i <= n; i++)
    {
        for(;;)
        {
            for(int j = 1; j <= m; j++) S[j] = T[j] = 0;
            if(match(i)) break; else update();
        }
    }
}

inline void readint(int &x)
{
    char c;
    c = getchar();
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c)) x = x*10+c-'0', c = getchar();
}

inline void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

int N, M;
int sum;

void read_case()
{
	n = N, m = M;
	
	for(int i = 1; i <= N; i++)
	for(int j = 1; j <= M; j++)
	{
		readint(W[i][j]);
		W[i][j] *= 100;
	}
	
	sum = 0;
	
	for(int i = 1; i <= N; i++)
	{
		int x; readint(x);
		sum += W[i][x];
		W[i][x] += 1;
	}
}

void solve()
{
	read_case();
	KM();
	
	int ans = 0;
	for(int i = 1; i <= m; i++) if(Left[i]) ans += W[Left[i]][i];
	
	int ans1 = N-ans%100, ans2 = ans/100 - sum/100;
	writeint(ans1), putchar(' '), writeint(ans2), puts("");
}

int main()
{
	while(~scanf("%d%d", &N, &M))
	{
		solve();
	}
	return 0;
}

/*HDU 3315*/
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cctype>
using namespace std;

const int maxn = 100;
const int INF = 0x3f3f3f3f;

int n, m;

int W[maxn][maxn];
int Lx[maxn], Ly[maxn];

int Left[maxn];
bool S[maxn], T[maxn];

bool match(int i)
{
    S[i] = 1;
    for(int j = 1; j <= m; j++) if(Lx[i]+Ly[j] == W[i][j] && !T[j])
    {
        T[j] = 1;
        if(!Left[j] || match(Left[j]))
        {
            Left[j] = i;
            return 1;
        }
    }
    return 0;
}

void update()
{
    int a = INF;
    for(int i = 1; i <= n; i++) if(S[i])
    for(int j = 1; j <= m; j++) if(!T[j])
        a = min(a, Lx[i]+Ly[j]-W[i][j]);
        
    for(int i = 1; i <= n; i++)
    {
        if(S[i]) Lx[i] -= a;
    }
    for(int j = 1; j <= m; j++)
    {
        if(T[j]) Ly[j] += a;
    }
}

void KM()
{
    memset(Left, 0, sizeof(Left));
    memset(Lx, 0, sizeof(Lx));
    memset(Ly, 0, sizeof(Ly));
    
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
            Lx[i] = max(Lx[i], W[i][j]);
    }
    for(int i = 1; i <= n; i++)
    {
        for(;;)
        {
            for(int j = 1; j <= m; j++) S[j] = T[j] = 0;
            if(match(i)) break; else update();
        }
    }
}

inline void readint(int &x)
{
    char c;
    c = getchar();
    while(!isdigit(c)) c = getchar();
    
    x = 0;
    while(isdigit(c)) x = x*10+c-'0', c = getchar();
}

inline void writeint(int x)
{
    if(x > 9) writeint(x/10);
    putchar(x%10+'0');
}

int N;
int V[maxn];
int H[maxn], P[maxn];
int A[maxn], B[maxn];

int check(int i, int j)
{
	int Hi = H[i], Pi = P[j];
	int d1 = A[i], d2 = B[j];
	for(;;)
	{
		Pi -= d1;
		if(Pi <= 0) return 1;
		Hi -= d2;
		if(Hi <= 0) return 0;
	}
	return -1;
}

void read_case()
{
	n = m = N;
	
	for(int i = 1; i <= N; i++) readint(V[i]);
	for(int i = 1; i <= N; i++) readint(H[i]);
	for(int i = 1; i <= N; i++) readint(P[i]);
	for(int i = 1; i <= N; i++) readint(A[i]);
	for(int i = 1; i <= N; i++) readint(B[i]);
}

void build()
{
	for(int i = 1; i <= N; i++)
	for(int j = 1; j <= N; j++)
	{
		if(check(i, j) > 0)
		{
			W[i][j] = V[i]*100;
		}
		else
		{
			W[i][j] = -V[i]*100;
		}
		if(i == j) W[i][j] += 1;
	}
}

void solve()
{
	read_case();
	build();
	
	KM();
	
	int ans = 0;
	for(int i = 1; i <= m; i++) if(Left[i]) ans += W[Left[i]][i];
	
	if(ans > 0)
	{
		int ans1 = ans/100;
		double ans2 = 1.0*(ans%100)/N * 100;
		printf("%d %.3lf%%\n", ans1, ans2);
	}
	else { printf("Oh, I lose my dear seaco!\n"); return ; }
}

int main()
{
	while(~scanf("%d", &N) && N)
	{
		solve();
	}
	return 0;
}