HDU 3315 KM My Brute

参考题解

二分图的最优匹配。图很容易建立。再处理相似度的时候。把每个权值扩大100倍。然后再对i==j时 特殊标记。使他们的权值再++1。后面选择的时候就很容易挑出。按原匹配

匹配的个数。 100*（double）(res%100)/n。即可得到第二问。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

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

int n;

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

// KM algorithm
int W[maxn][maxn], lft[maxn];
int slack[maxn];
int Lx[maxn], Ly[maxn];
bool S[maxn], T[maxn];

int inline divide(int a, int b) { int ans = a / b; if(a % b) ans++; return ans; }

bool match(int u)
{
    S[u] = true;
    for(int v = 1; v <= n; v++) if(!T[v])
    {
        int t = Lx[u] + Ly[v] - W[u][v];
        if(t == 0)
        {
            T[v] = true;
            if(!lft[v] || match(lft[v]))
            {
                lft[v] = u;
                return true;
            }
        }
        else slack[v] = min(slack[v], t);
    }

    return false;
}

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

        if(T[i]) Ly[i] += a;
        else slack[i] -= a;
    }
}

void KM()
{
    memset(Ly, 0, sizeof(Ly));
    memset(lft, 0, sizeof(lft));
    for(int i = 1; i <= n; i++) Lx[i] = -INF;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++) Lx[i] = max(Lx[i], W[i][j]);

    for(int i = 1; i <= n; i++)
    {
        memset(slack, 0x3f, sizeof(slack));
        for(;;)
        {
            memset(S, false, sizeof(S));
            memset(T, false, sizeof(T));
            if(match(i)) break;
            update();
        }
    }
}

int main()
{
    while(scanf("%d", &n) == 1 && n)
    {
        for(int i = 1; i <= n; i++) scanf("%d", V + i);
        for(int i = 1; i <= n; i++) scanf("%d", H + i);
        for(int i = 1; i <= n; i++) scanf("%d", P + i);
        for(int i = 1; i <= n; i++) scanf("%d", A + i);
        for(int i = 1; i <= n; i++) scanf("%d", B + i);

        //build graph
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
            {
                int score = V[i] * 100;
                int round1 = divide(P[j], A[i]);
                int round2 = divide(H[i], B[j]);
                if(round1 > round2) score = -score;
                if(i == j) score++;
                W[i][j] = score;
            }

        KM();

        int ans = 0;
        for(int i = 1; i <= n; i++) ans += W[lft[i]][i];

        if(ans < 0)
        {
            puts("Oh, I lose my dear seaco!");
            continue;
        }

        printf("%d ", ans / 100);
        printf("%.3f%%
", 100.0 * (ans % 100) / n);
    }

    return 0;
}