[USACO] OPEN11 Bronze Division

PROBLEM 11: bfire

題目說了一大堆，其實抽象成編程問題就是：給定N個數1～n放在編號爲1~n的位置上，現在進行移動使得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越界則從位置1接着計數，依次類推，問哪個數會出現在1號位置或者哪個數其將要放的位置上的數是已經被移動過的了？舉例來說：

1 2 3 -> 2

1 2 3 4 -> 4

1 2 3 4 5 -> 3

1 2 3 4 5 6 -> 4

解法其實很簡單，模擬一下數字的移動過程就行了，邊界條件就是上述的那兩個停止條件。

 

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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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注意註釋部分，爲了方便，我的數組下標是從1開始的，這樣所有2的k次冪，比如128，求得的結果爲0，其實應該是128，所以不應該取模，直接減去一個n就可以了。

 

<------------------------------------------------------------------------瀟灑的分割線------------------------------------------------------------------------------------------->

 

PROBLEM 12: ssort

有點像快排，非常清晰的遞歸求解，測試全過，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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<------------------------------------------------------------------------幽怨的分割線------------------------------------------------------------------------------------------->

 

PROBLEM 13: space3d

看了一下分析，提到了一個Flood Fill算法，上網搜了搜，寫了一個遞歸的解決方案，結果segmentation fault。。。

想了一下應該是棧溢出了，遞歸的層次太深了

參考這篇文章《Flood-Fill 種子填充》 

/* 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); }  

繼續優化，參考了這篇文章《種子填充的優化》

也沒怎麼仔細看，就看個思路，利用隊列解決遞歸的低效，下面是代碼

/* 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); }

怎麼說呢，首先程序寫的相當爛！那六個if語句真是爛透了！！看看別人的程序就知道了。。。（這個以後會改進）

其次，讓我耗了半個下午的地方就是內存分配超出限制，queue數組和visit數組都聲明爲int類型時，程序超出內存限制

看了別人的解法嘗試着把這兩個數組聲明爲char類型，就過了，無語。。。

char visit -> 15636KB

char queue -> 9784KB

char both -> 6852KB

 

 

 

<------------------------------------------------------------------------蛋疼的分割線------------------------------------------------------------------------------------------->

 

PROBLEM 14: stringe

題目很簡單，不過竟然segmentation fault！！原來定義緩衝區長度定義小了，比如

0 9 CHESS這個數據，其實執行玩之後應該有2的9次冪個CHESS，所以需要修正宏定義：

#define MAX_LEN LEN * 12 改成

#define MAX_LEN LEN * 1<<12

最大的開銷是需要分配兩個400多K的靜態數組。。。

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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結果你妹的超時了！繼續優化，去掉結構體，串長賦值給一個變量，兩個緩衝區交替輸出結果，避免不必要的字符串拷貝，很像圖像處理中的後備緩衝區的用法。。。

/* 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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此題再一次印證了“程序不是調試出來的”這一警句，寫程序時就應該注意到空間分配，程序效率等問題，良好的設計才能產出良好的結果，程序就是要追求簡單，高效，像你前一個程序，冗餘的字符串拷貝難道不需要時間嗎？擦！