编程之美2014格格取数
2014-04-12 13:02
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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
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 */
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