HDU 3395 Special Fish

HDU_3395

    我们可以把一条鱼拆成两个点分别代表攻击和被攻击，然后按题意对G[i][j]==1的边的权值设为value[i] ^ value[j]，之后用KM算法求二分图的最优匹配即可。

#include<stdio.h>
#include<string.h>
#define MAXD 110
#define INF 1000000000
int yM[MAXD], G[MAXD][MAXD], N, value[MAXD];
int A[MAXD], B[MAXD];
int visx[MAXD], visy[MAXD], slack;
char b[MAXD];
int init()
{
    int i, j;
    scanf("%d", &N);
    if(!N)
        return 0;
    for(i = 0; i < N; i ++)
        scanf("%d", &value[i]);
    for(i = 0; i < N; i ++)
    {
        scanf("%s", b);
        for(j = 0; j < N; j ++)
        {
            G[i][j] = b[j] - '0';
            if(G[i][j])
                G[i][j] = (value[i] ^ value[j]);
        }
    }
    return 1;
}
int searchpath(int u)
{
    int v, temp;
    visx[u] = 1;
    for(v = 0; v < N; v ++)
        if(!visy[v])
        {
            temp = A[u] + B[v] - G[u][v];
            if(temp == 0)
            {
                visy[v] = 1;
                if(yM[v] == -1 || searchpath(yM[v]))
                {
                    yM[v] = u;
                    return 1;
                }
            }
            else if(temp < slack)
                slack = temp;
        }
    return 0;
}
void KM()
{
    int i, j, u;
    for(i = 0; i < N; i ++)
    {
        A[i] = 0;
        for(j = 0; j < N; j ++)
            if(G[i][j] > A[i])
                A[i] = G[i][j];
    }
    memset(B, 0, sizeof(B));
    memset(yM, -1, sizeof(yM));
    for(u = 0; u < N; u ++)
        for(;;)
        {
            memset(visx, 0, sizeof(visx));
            memset(visy, 0, sizeof(visy));
            slack = INF;
            if(searchpath(u))
                break;
            for(i = 0; i < N; i ++)
            {
                if(visx[i])
                    A[i] -= slack;
                if(visy[i])
                    B[i] += slack;
            }
        }
}
void printresult()
{
    int i, res = 0;
    for(i = 0; i < N; i ++)
        res += G[yM[i]][i];
    printf("%d\n", res);
}
int main()
{
    while(init())
    {
        KM();
        printresult();
    }
    return 0;
}