编程之美2014格格取数

最新推荐文章于 2015-04-12 21:03:58 发布

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最新推荐文章于 2015-04-12 21:03:58 发布

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文章标签： 2014编程之美

本文链接：https://blog.youkuaiyun.com/littlefool5201314/article/details/23525391

本文介绍了使用匈牙利算法思想解决编程之美2014年竞赛中的一道题目，作者在尝试过程中遇到了AC与WA的情况，分享了代码和疑惑。

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根据匈牙利算法解决的思想改写的代码

此题我觉得题意跟数据的有冲突

3 3

1 2 3
3 2 1
2 2 2

上面那段结果是6的能AC 结果是4的是WA 不晓得是怎么回事。

下面这段代码是根据我的思路写的，但是没有AC

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
using namespace std;
#define Max 107
int n,n1;
int s[Max][Max];
int p[Max][Max];
int q[Max][Max];
int row[Max], col[Max];
int c1[Max],c2[Max];
int r[Max][Max];
int x[Max],y[Max];
void countZero()
{
    memset(row, 0, sizeof(row));
    memset(col, 0, sizeof(col));
    memset(r, 0, sizeof(r));
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < n1; ++j)
        {
            if (p[i][j] == 0)
                row[i]++, col[j]++;
        }
    }
}

int drawLine()
{
    memset(q, 0, sizeof(q));
    for (int i = 0; i < n; ++i)
        x[i] = 1;
    for(int j=0;j<n1;j++)
         y[j] = 0;
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < n1; ++j)
        {
            if (r[i][j] == 2)
            {
                x[i] = 0;
                break;
            }
        }
    }
    bool is = 1;
    while (is)
    {
        is = 0;
        for (int i = 0; i < n; ++i)
        {
            if (x[i] == 1)
            {
                for (int j = 0; j < n1; ++j)
                {
                    if (p[i][j] == 0 && y[j] == 0)
                    {
                        y[j] = 1;
                        is = 1;
                    }
                }
            }
        }
        for (int j = 0; j < n1; ++j)
        {
            if (y[j] == 1)
            {
                for (int i = 0; i < n; ++i)
                {
                    if (p[i][j] == 0 && x[i] == 0 && r[i][j] == 2)
                    {
                        x[i] = 1;
                        is = 1;
                    }
                }
            }
        }
    }
    int line = 0;
    for (int i = 0; i < n; ++i)
    {
        if (x[i] == 0)
        {
            for (int j = 0; j < n1; ++j)
                q[i][j]++;
            line++;
        }
    }
    for(int i=0;i<n1;i++)
    {
        if (y[i] == 1)
        {
            for (int j = 0; j < n; ++j)
                q[j][i]++;
            line++;
        }
    }
    return line;
}

int find()
{
    countZero();
    int zero = 0;
    while (1)
    {
        for (int i = 0; i < n; ++i)
        {
            if (row[i] == 0)
                row[i] = INT_MAX;
        }
        for(int i=0; i < n1 ;i++)
        {
            if (col[i] == 0)
                col[i] = INT_MAX;
        }
        bool stop = 1;
        if (*min_element(row, row+n) <= *min_element(col, col+n1))
        {
            int tmp = INT_MAX, index = -1;
            for (int i = 0; i < n; ++i)
            {
                if (tmp > row[i])
                    tmp = row[i], index = i;
            }
            int index2 = -1;
            for (int i = 0; i < n1; ++i)
                if (p[index][i] == 0 && col[i] != INT_MAX)
                {
                    index2 = i;
                    stop = 0;
                    zero++;
                    break;
                }
            if (stop)
                break;
            row[index] = col[index2] = INT_MAX;
            r[index][index2] = 1;
            for (int i = 0; i < n1; ++i)
                if (p[index][i] == 0 && col[i] != INT_MAX)
                    col[i]--;
            for (int i = 0; i < n; ++i)
                if (p[i][index2] == 0 && row[i] != INT_MAX)
                    row[i]--;
        }
        else
        {
            int tmp = INT_MAX, index = -1;
            for (int i = 0; i < n1; ++i)
            {
                if (tmp > col[i])
                    tmp = col[i], index = i;
            }
            int index2 = -1;
            for (int i = 0; i < n; ++i)
                if (p[i][index] == 0 && row[i] != INT_MAX)
                {
                    index2 = i;
                    stop = 0;
                    zero++;
                    break;
                }

            if (stop)
                break;
            row[index2] = col[index] = INT_MAX;
            r[index2][index] = 1;

            for (int i = 0; i < n1; ++i)
                if (p[index2][i] == 0 && col[i] != INT_MAX)
                    col[i]--;
            for (int i = 0; i < n; ++i)
                if (p[i][index] == 0 && row[i] != INT_MAX)
                    row[i]--;
        }
    }
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n1; ++j)
            if (p[i][j] == 0)
                r[i][j]++;

    return zero;
}

int main()
{
    int m,l;
    scanf("%d",&m);
    for(l=1;l<=m;l++)
    {
        int ans = 0;
        scanf("%d%d",&n,&n1);
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n1; ++j)
                scanf("%d",&s[i][j]);
      //  printf("求和最小\n");
        for (int i = 0; i < n; ++i)
            {for (int j = 0; j < n1; ++j)
              //  p[i][j] = *max_element(s[i], s[i]+n)-s[i][j],    //求和最大
                p[i][j] = s[i][j]-*min_element(s[i],s[i]+n1);
                //printf("%d ",p[i][j]);
               // printf("\n");
                }     //求和最小
        for (int j = 0; j < n1; ++j)
        {
            int tmp = INT_MAX;
            for (int i = 0; i < n; ++i)
            {
                if (tmp > p[i][j])
                    tmp = p[i][j];
            }
            for (int i = 0; i < n; ++i)
                p[i][j] -= tmp;
        }
       /* printf("第一次减小最小值\n");
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < n1; ++j)
                printf("%d ",p[i][j]);
            printf("\n");
        }
    */
        while (find() < n&&find()<n1)
        {
            drawLine();
            int min = INT_MAX;
            for (int i = 0; i < n; ++i)
                for (int j = 0; j < n1; ++j)
                    if (q[i][j] == 0 && min > p[i][j])
                        min = p[i][j];

            for (int i = 0; i < n; ++i)
                for (int j = 0; j < n1; ++j)
                    if (q[i][j] == 0)
                        p[i][j] -= min;
                    else if (q[i][j] == 2)
                        p[i][j] += min;
        }
    //    printf("输出划线后的状态\n");
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        for (int i = 0; i < n; ++i)
            {for (int j = 0; j < n1; ++j)
                {
                  //  printf("%d ",r[i][j]);
                    if (r[i][j] == 2)
                   {ans += s[i][j];c1[i]=1;c2[j]=1;}
                }
        //        printf("\n");
            }
        if(n>n1)
        {
            for(int i=0;i<n;i++)
            {
                if(!c1[i])
                {
                    ans+=*min_element(s[i],s[i]+n1);
                }
            }
        }
        if(n<n1)
        {
            for(int j=0;j<n1;j++)
            {
                int min2=s[0][j];
                if(!c2[j])
                {
                    for(int i=0;i<n;i++)
                    {
                        if(min2>s[i][j]) min2=s[i][j];
                    }
                    ans+=min2;
                }
            }
        }
        printf("Case %d: %d\n",l,ans);
    }
    return 0;
}
/* 一些输出样例
333
3 3
3 2 1
1 2 3
2 2 2
Case 1: 4
3 1
3
2
1
Case 2: 6
1 3
1 2 3
Case 3: 6
3 3
0 0 0
0 2 1
0 0 0
Case 4: 0
*/