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
- 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 105using 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 |
