很直接的一道题，排序比较就行了，不过还是超时了一次，因为一不小心写成死循环了。。。
代码如下:
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 5 const int maxn = 20000+10; 6 int h[maxn], k[maxn]; 7 8 int main() 9 { 10 #ifdef LOCAL 11 freopen("in", "r", stdin); 12 #endif 13 int n, m; 14 while(scanf("%d%d", &n, &m) != EOF && n && m) 15 { 16 for(int i = 0; i < n; i++) 17 scanf("%d", &h[i]); 18 for(int i = 0; i < m; i++) 19 scanf("%d", &k[i]); 20 sort(h, h+n); 21 sort(k, k+m); 22 int i, j = 0; 23 int flag = 1; 24 int sum = 0; 25 for(i = 0; i < n; i++) 26 { 27 while(1) 28 { 29 if(j >= m) 30 { 31 flag = 0; 32 break; 33 } 34 if(k[j] >= h[i]) 35 { 36 sum += k[j]; 37 j++; 38 break; 39 } 40 j++; 41 } 42 if(!flag) break; 43 } 44 if(!flag) printf("Loowater is doomed!\n"); 45 else printf("%d\n", sum); 46 } 47 return 0; 48 }
这是开始写的代码，把头作外层循环，让每一个骑士去找头，不仅不合正常思维，而且代码长、难理解，后来看了下书，修改代码如下：
1 #include <cstdio> 2 #include <algorithm> 3 #define MAXN 20000+10 4 using namespace std; 5 6 int main() 7 { 8 #ifdef LOCAL 9 freopen("in", "r", stdin); 10 #endif 11 int h[MAXN], k[MAXN]; //h[] are dragon's heads and k[] are knights 12 int n, m; // n is the number of heads, m is the number of knights 13 while (scanf("%d%d", &n, &m) != EOF) 14 { 15 if (n == 0 && m == 0) break; 16 for (int i = 0; i < n; i++) scanf("%d", &h[i]); 17 for (int i = 0; i < m; i++) scanf("%d", &k[i]); 18 sort(h, h+n); 19 sort(k, k+m); 20 int cur = 0, cost = 0; 21 for (int i = 0; i < m; i++) // for each knight 22 if (k[i] >= h[cur]) 23 { 24 cost += k[i]; 25 if (++cur >= n) break; 26 } 27 if (cur < n) printf("Loowater is doomed!\n"); 28 else printf("%d\n", cost); 29 } 30 return 0; 31 }
这下感觉不错，自然，代码也短了。然后就像看看c能比c++快多少，就又改了一下，变成c版本：
1 #include <stdio.h> 2 #include <stdlib.h> 3 #define MAXN 20000+10 4 5 int cmp(const void *_a, const void *_b) 6 { 7 int *a = (int*)_a; 8 int *b = (int*)_b; 9 return *a - *b; 10 } 11 12 int main() 13 { 14 #ifdef LOCAL 15 freopen("in", "r", stdin); 16 #endif 17 int h[MAXN], k[MAXN]; /* h[] are dragon's heads and k[] are knights */ 18 int n, m, i; /* n is the number of heads, m is the number of knights */ 19 int cur, cost; 20 while (scanf("%d%d", &n, &m) != EOF) 21 { 22 if (n == 0 && m == 0) break; 23 for (i = 0; i < n; i++) scanf("%d", &h[i]); 24 for (i = 0; i < m; i++) scanf("%d", &k[i]); 25 qsort(h, n, sizeof(int), cmp); 26 qsort(k, m, sizeof(int), cmp); 27 cur = 0; 28 cost = 0; 29 for (i = 0; i < m; i++) /* for each knight */ 30 if (k[i] >= h[cur]) 31 { 32 cost += k[i]; 33 if (++cur >= n) break; 34 } 35 if (cur < n) printf("Loowater is doomed!\n"); 36 else printf("%d\n", cost); 37 } 38 return 0; 39 }
结果...c++的代码用了29ms，c的代码用了32ms，有点惊讶...不过还是喜欢c++的随处声明变量的方式，强大的STL函数库、还有习惯了的注释（//注释风格在gcc的ansi选项会出错），同时还不习惯c++式的输入输出，暂时还是用c的吧，等我发现c++输入输出的好处再改啦(*^__^*) 嘻嘻……