[USACO] OPEN11 Bronze Division

PROBLEM 11: bfire

题目说了一大堆，其实抽象成编程问题就是：给定N个数1～n放在编号为1~n的位置上，现在进行移动使得n1放到n2的位置上，n2放到n4的位置上，nk放到n2k的位置上，如果n2k越界则从位置1接着计数，依次类推，问哪个数会出现在1号位置或者哪个数其将要放的位置上的数是已经被移动过的了？举例来说：

1 2 3 -> 2

1 2 3 4 -> 4

1 2 3 4 5 -> 3

1 2 3 4 5 6 -> 4

解法其实很简单，模拟一下数字的移动过程就行了，边界条件就是上述的那两个停止条件。

 

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/* ID: LANG: C TASK: bfire */ #include <stdio.h> #include <string.h> #include <stdlib.h> #define N 250 int main() { FILE *fin = fopen("bfire.in", "r"); FILE *fout = fopen("bfire.out", "w"); int cows[N+1] = {0}; int n, cur, next; fscanf(fin, "%d", &n); next = 2; while(next != 1 && cows[next] == 0) { cows[next] = 1; cur = next; next += next; if(next > n) next = next - n;//next = next % n; } fprintf(fout, "%d/n", cur); exit(0); }  

 

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注意注释部分，为了方便，我的数组下标是从1开始的，这样所有2的k次幂，比如128，求得的结果为0，其实应该是128，所以不应该取模，直接减去一个n就可以了。

 

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PROBLEM 12: ssort

有点像快排，非常清晰的递归求解，测试全过，爽！

 

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/* ID: LANG: C TASK: ssort */ #include <stdio.h> #include <stdlib.h> #define N 10 int dist_moved = 0; void ssort(int *cows, int start, int size) { if(size == 1) return; int half = size / 2; int i, j, tmp; ssort(cows, start, half); ssort(cows, start+half, half); if(cows[start] > cows[start + half]) { j = start + half; for(i = 0; i < half; i++) { tmp = cows[start + i]; cows[start + i] = cows[j + i]; cows[j + i] = tmp; } dist_moved += half * size; } } int main() { FILE *fin = fopen("ssort.in", "r"); FILE *fout = fopen("ssort.out", "w"); int cows[1<<N]; int i, j, n; fscanf(fin, "%d", &n); for(i = 0; i < 1<<n; i++) { fscanf(fin, "%d", &j); cows[i] = j; } ssort(cows, 0, 1<<n); fprintf(fout, "%d/n", dist_moved); for(i = 0; i < 1<<n; i++) fprintf(fout, "%d/n", cows[i]); exit(0); }  

 

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PROBLEM 13: space3d

看了一下分析，提到了一个Flood Fill算法，上网搜了搜，写了一个递归的解决方案，结果segmentation fault。。。

想了一下应该是栈溢出了，递归的层次太深了

参考这篇文章《Flood-Fill 种子填充》 

/* ID: LANG: C TASK: space3d */ #include <stdio.h> #include <stdlib.h> #define N 100 void flood_fill(char (*s)[N][N], int (*v)[N][N], int n, int i, int j, int k) { if(i<0 || j<0 || k<0 || i>n || j>n || k>n || v[i][j][k] || s[i][j][k] != '*') return; v[i][j][k] = 1; //printf("(%d, %d, %d) is visited/n", i, j, k); flood_fill(s, v, n, i-1, j, k); flood_fill(s, v, n, i+1, j, k); flood_fill(s, v, n, i, j-1, k); flood_fill(s, v, n, i, j+1, k); flood_fill(s, v, n, i, j, k-1); flood_fill(s, v, n, i, j, k+1); } int main() { FILE *fin = fopen("space3d.in", "r"); FILE *fout = fopen("space3d.out", "w"); char c, space[N][N][N]; int visit[N][N][N]; int n, count = 0; int i, j, k; fscanf(fin, "%d", &n); getc(fin); for(k = 0; k < n; k++) for(i = 0; i < n; i++) { for(j = 0; j < n; j++) { c = getc(fin); space[i][j][k] = c; visit[i][j][k] = 0; } getc(fin); } for(k = 0; k < n; k++) for(i = 0; i < n; i++) for(j = 0; j < n; j++) if(space[i][j][k] == '*' && !visit[i][j][k]) { flood_fill(space, visit, n, i, j, k); count++; } fprintf(fout, "%d/n", count); exit(0); }  

继续优化，参考了这篇文章《种子填充的优化》

也没怎么仔细看，就看个思路，利用队列解决递归的低效，下面是代码

/* ID: LANG: C TASK: space3d */ #include <stdio.h> #include <stdlib.h> #define N 100 //int queue[N*N*N+1][3]; char queue[N*N*N+1][3]; char space[N][N][N]; char visit[N][N][N]; void flood_fill(char (*s)[N][N], char (*v)[N][N], int n, int i, int j, int k) { //char queue[N*N*N+1][3]; int head, end; head = end = 0; v[i][j][k] = 1; queue[end][0] = i; queue[end][1] = j; queue[end][2] = k; end++; //printf("add (%d, %d, %d), head = %d, end = %d/n", i, j, k, head, end); while(head < end) { i = queue[head][0]; j = queue[head][1]; k = queue[head][2]; head++; if(i+1>=0 && i+1<n && s[i+1][j][k] == '*' && !v[i+1][j][k]) { v[i+1][j][k] = 1; queue[end][0] = i+1; queue[end][1] = j; queue[end][2] = k; end++; //printf("add (%d, %d, %d), head = %d, end = %d/n", i+1, j, k, head, end); } if(i-1>=0 && i-1<n && s[i-1][j][k] == '*' && !v[i-1][j][k]) { v[i-1][j][k] = 1; queue[end][0] = i-1; queue[end][1] = j; queue[end][2] = k; end++; //printf("add (%d, %d, %d), head = %d, end = %d/n", i-1, j, k, head, end); } if(j+1>=0 && j+1<n && s[i][j+1][k] == '*' && !v[i][j+1][k]) { v[i][j+1][k] = 1; queue[end][0] = i; queue[end][1] = j+1; queue[end][2] = k; end++; //printf("add (%d, %d, %d), head = %d, end = %d/n", i, j+1, k, head, end); } if(j-1>=0 && j-1<n && s[i][j-1][k] == '*' && !v[i][j-1][k]) { v[i][j-1][k] = 1; queue[end][0] = i; queue[end][1] = j-1; queue[end][2] = k; end++; //printf("add (%d, %d, %d), head = %d, end = %d/n", i, j-1, k, head, end); } if(k+1>=0 && k+1<n && s[i][j][k+1] == '*' && !v[i][j][k+1]) { v[i][j][k+1] = 1; queue[end][0] = i; queue[end][1] = j; queue[end][2] = k+1; end++; //printf("add (%d, %d, %d), head = %d, end = %d/n", i, j, k+1, head, end); } if(k-1>=0 && k-1<n && s[i][j][k-1] == '*' && !v[i][j][k-1]) { v[i][j][k-1] = 1; queue[end][0] = i; queue[end][1] = j; queue[end][2] = k-1; end++; //printf("add (%d, %d, %d), head = %d, end = %d/n", i, j, k-1, head, end); } } } int main() { FILE *fin = fopen("space3d.in", "r"); FILE *fout = fopen("space3d.out", "w"); char c; //char space[N][N][N]; //int visit[N][N][N]; int n, count = 0; int i, j, k; fscanf(fin, "%d", &n); getc(fin); for(k = 0; k < n; k++) for(i = 0; i < n; i++) { for(j = 0; j < n; j++) { c = getc(fin); space[i][j][k] = c; visit[i][j][k] = 0; } getc(fin); } for(k = 0; k < n; k++) for(i = 0; i < n; i++) for(j = 0; j < n; j++) if(space[i][j][k] == '*' && !visit[i][j][k]) { //printf("flood center (%d, %d, %d)/n", i, j, k); flood_fill(space, visit, n, i, j, k); count++; } fprintf(fout, "%d/n", count); //printf("%d/n", count); exit(0); }

怎么说呢，首先程序写的相当烂！那六个if语句真是烂透了！！看看别人的程序就知道了。。。（这个以后会改进）

其次，让我耗了半个下午的地方就是内存分配超出限制，queue数组和visit数组都声明为int类型时，程序超出内存限制

看了别人的解法尝试着把这两个数组声明为char类型，就过了，无语。。。

char visit -> 15636KB

char queue -> 9784KB

char both -> 6852KB

 

 

 

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PROBLEM 14: stringe

题目很简单，不过竟然segmentation fault！！原来定义缓冲区长度定义小了，比如

0 9 CHESS这个数据，其实执行玩之后应该有2的9次幂个CHESS，所以需要修正宏定义：

#define MAX_LEN LEN * 12 改成

#define MAX_LEN LEN * 1<<12

最大的开销是需要分配两个400多K的静态数组。。。

下面是程序

 

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/* ID: LANG: C TASK: stringe */ #include <stdio.h> #include <stdlib.h> #include <string.h> #define Z 100 #define LEN 100 #define MAX_LEN LEN * 1<<12 typedef struct { char s[LEN+1]; int N_i, C_i; }stringe; int main() { FILE *fin = fopen("stringe.in", "r"); FILE *fout = fopen("stringe.out", "w"); stringe str[Z]; char sbuf[MAX_LEN+1], tbuf[MAX_LEN+1]; int i, j, k, n; fscanf(fin, "%d", &n); for(i = 0; i < n; i++) fscanf(fin, "%d %d %s", &str[i].N_i, &str[i].C_i, str[i].s); for(i = 0; i < n; i++) { strcpy(sbuf, str[i].s); for(j = 0; j < str[i].C_i; j++) { memset(tbuf, '/0', MAX_LEN); for(k = str[i].N_i; k < strlen(sbuf); k++) tbuf[k - str[i].N_i] = sbuf[k]; strcat(tbuf, sbuf); strcpy(sbuf, tbuf); } fprintf(fout, "%s/n", tbuf); //printf("%s/n/n", tbuf); } exit(0); }  

 

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结果你妹的超时了！继续优化，去掉结构体，串长赋值给一个变量，两个缓冲区交替输出结果，避免不必要的字符串拷贝，很像图像处理中的后备缓冲区的用法。。。

/* ID: LANG: C TASK: stringe */ #include <stdio.h> #include <stdlib.h> #include <string.h> #define LEN 100 #define MAX_LEN LEN * 1<<12 int main() { FILE *fin = fopen("stringe.in", "r"); FILE *fout = fopen("stringe.out", "w"); char sbuf[MAX_LEN+1], tbuf[MAX_LEN+1]; int i, j, k, n, len; int N_i, C_i; fscanf(fin, "%d", &n); for(i = 0; i < n; i++) { fscanf(fin, "%d %d %s", &N_i, &C_i, sbuf); for(j = 0; j < C_i; j++) { if(j % 2 == 0) { memset(tbuf, '/0', MAX_LEN); len = strlen(sbuf); for(k = N_i; k < len; k++) tbuf[k - N_i] = sbuf[k]; strcat(tbuf, sbuf); } else{ memset(sbuf, '/0', MAX_LEN); len = strlen(tbuf); for(k = N_i; k < len; k++) sbuf[k - N_i] = tbuf[k]; strcat(sbuf, tbuf); } } if(j % 2 == 0) fprintf(fout, "%s/n", sbuf); else fprintf(fout, "%s/n", tbuf); } exit(0); }  

 

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此题再一次印证了“程序不是调试出来的”这一警句，写程序时就应该注意到空间分配，程序效率等问题，良好的设计才能产出良好的结果，程序就是要追求简单，高效，像你前一个程序，冗余的字符串拷贝难道不需要时间吗？擦！