編程之美2014格格取數

根據匈牙利演算法解決的思想改寫的代碼

此題我覺得題意跟資料的有衝突

3 3

1 2 3
3 2 1
2 2 2

上面那段 結果是6的能AC 結果是4的是WA 不曉得是怎麼回事。

下面這段代碼是根據我的思路寫的，但是沒有AC

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <climits>using namespace std;#define Max 107int 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("第一次減小最小值\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("輸出劃線後的狀態\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;}/* 一些輸出範例3333 33 2 11 2 32 2 2Case 1: 43 1321Case 2: 61 31 2 3Case 3: 63 30 0 00 2 10 0 0Case 4: 0*/