题目1102：最小面积子矩阵

题目1102：最小面积子矩阵

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：510

解决：113

题目描述：

一个N*M的矩阵，找出这个矩阵中所有元素的和不小于K的面积最小的子矩阵（矩阵中元素个数为矩阵面积）

输入：

每个案例第一行三个正整数N,M<=100，表示矩阵大小，和一个整数K
接下来N行，每行M个数，表示矩阵每个元素的值

输出：

输出最小面积的值。如果出现任意矩阵的和都小于K，直接输出-1。

样例输入：

4 4 10
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

样例输出：

1

来源：

2010年上海交通大学计算机研究生机试真题

答疑：

解题遇到问题?分享解题心得?讨论本题请访问： http://t.jobdu.com/thread-7825-1-1.html

#include <stdio.h>
 
const int size = 110;
 
int mat[size][size];
int sum[size][size];
int tok[size];
int n, m, k;
 
bool check(int len)
{
    for (int i = 1; i + len - 1 <= m; i ++)
        if (tok[i + len - 1] - tok[i - 1] >= k)
            return true;
    return false;
}
 
int main()
{
    int a;
    while (scanf("%d %d %d", &n, &m, &k) != EOF)
    {
        for (int i = 1; i <= m; i ++)
            sum[0][i] = 0;
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= m; j ++)
            {
                scanf("%d", &mat[i][j]);
                sum[i][j] = sum[i-1][j] + mat[i][j];
            }
 
        int ans = -1;
        for (int len = 1; len <= n; len ++)
            for (int i = 1; i + len - 1 <= n; i ++)
            {
                tok[0] = 0;
                for (int j = 1; j <= m; j ++)
                {
                    tok[j] = sum[i + len - 1][j] - sum[i - 1][j];
                    tok[j] += tok[j-1];
                }
                if (tok[m] < k)
                  continue;
              //  printf("len = %d, %d!!\n", len, tok[m]);
                //枚举len，二分mid
                int lf = 1, rt = m + 1;
                int mid;
                while (lf < rt)
                {
                    mid = (lf + rt) >> 1;
                 //  printf("....(%d, %d)\n", lf, rt);
                    if (ans != -1 && mid * len > ans)
                    {
                        rt = mid;
                        continue;
                    }
                    if (check(mid))
                    {
                        if (ans == -1 || ans > mid * len)
                            ans = mid * len;
                        rt = mid;
                       //printf("ans = %d\n", ans);
                    }
                    else lf = mid + 1;
                }
            }
        printf("%d\n", ans);
    }
    return 0;
}
/*
1 5 10
1 2 5 3 7
*/
 
/**************************************************************
    Problem: 1102
    User: scode
    Language: C++
    Result: Accepted
    Time:10 ms
    Memory:1108 kb
**************************

我的 超时

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cstring>
#define min(a,b) ((a)>(b)?(b):(a))
#define max(a,b) ((a)<(b)?(b):(a))
int a[101][101];
int main()
{
//freopen("in.txt","r",stdin);
int n,m,k,i,j,p,q,temp,min,size1,size2,sum,res; bool flag;
while (scanf("%d %d %d", &n,&m,&k)!=EOF)
{
memset(a,0,sizeof(a));flag=false;res=100000;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&a[i][j]);
for(size1=1;size1<=n;size1++)//size
{
for(size2=1;size2<=m;size2++)
{
//if(flag==true) break;

for(i=1;i<=n-size1+1;i++)
{
for(j=1;j<=m-size2+1;j++)
{
sum=0;
for(p=0;p<size1;p++)
for(q=0;q<size2;q++)
sum+=a[i+p][j+q];
if( sum>=k ) { flag=true; if(res>size1*size2) res=size1*size2; }

//if(size1==3&&size2==1) printf("in&&&i %d j %d size1 %dsize2 %d sum %d\n",i,j,size1,size2,sum);
}
//if(size1==3&&size2==1) printf("i %d j %d size1 %d size2 %d sum %d\n",i,j,size1,size2,sum);
}
}
}
if(flag) printf("%d\n",res);
else printf("-1\n");

}

return 0;
}

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124

#include <cstdlib> #include <cstdio> #include <cstring> #include <algorithm> #define MAXN 105 using namespace std;   int N, M, K, G[MAXN][MAXN], S[MAXN*MAXN], lim; int rec[MAXN*MAXN], idx, sum[MAXN][MAXN];   struct Node {      int x, y; }que[MAXN];   int front, tail;   inline void pre() {      int k;      lim = N*M;      memset (S, 0, sizeof (S));      idx = -1;      // getstatus      for ( int i = 1; i <= N; ++i) {          k = min(M, lim / i); // 选取相对小的值,排除不合法情况          for ( int j = 1; j <= k; ++j) {              S[i*j] = 1;          }      }      for ( int i = 1; i <= lim; ++i) {          if (S[i]) {              rec[++idx] = i;          }      }      // getsum      for ( int i = 1; i <= N; ++i) {          for ( int j = 1; j <= M; ++j) {              sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + G[i][j];          }      } }   bool judge() {      int xx, yy, zz, lx, ly;      for ( int k = 1; k <= tail; ++k) {          lx = N - que[k].x + 1, ly = M - que[k].y + 1;          for ( int i = 1; i <= lx; ++i) {              for ( int j = 1; j <= ly; ++j) {                  xx = i+que[k].x-1, yy = j+que[k].y-1;                  if (xx <= N && yy <= M) {                      zz = sum[xx][yy] - sum[i-1][yy] - sum[xx][j-1] + sum[i-1][j-1];                      if (zz >= K) return true ;                  }              }          }      }      return false ; }   bool Ac( int x) // 能够保证所有的x都是合法值 {      front = tail = 0;      for ( int i = 1; i <= N; ++i) {          if (x % i == 0) {              ++tail;              que[tail].x = i, que[tail].y = x / i;          }      } // 处理处当前面积的分解情况      if (judge()) {          return true ;      }      return false ; }   int bsearch ( int l, int r) {      int mid, ret = -1;      while (l <= r) {          mid = (l + r) >> 1;          if (Ac(rec[mid])) {              ret = rec[mid];              r = mid - 1;          }          else {              l = mid + 1;             }      }      return ret; }   int force() {      for ( int i = 0; i <= idx; ++i) {          if (Ac(rec[i])) {              return rec[i];          }         }      return -1; }   int main() {      while ( scanf ( "%d %d %d" , &N, &M, &K) != EOF) {          for ( int i = 1; i <= N; ++i) {              for ( int j = 1; j <= M; ++j) {                  scanf ( "%d" , &G[i][j]);              }          }          pre();      //    printf("%d\n", bsearch(0, idx));          printf ( "%d\n" , force());      }      return 0;    } /**************************************************************      Problem: 1102      User: huashiyiqike      Language: C++      Result: Accepted      Time:40 ms      Memory:1184 kb

 

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