贪心算法。先用vis数组标记此栏里是否有牛(vis[i]==1表示第i栏有牛),然后用数组gap记录从此栏开始的最大空缺(比如X表示有牛,O表示没牛:XXOOOOXOX,编号从1开始。那么在这种情况下gap[2]==4,gap[7]==1,其他位置gap值均为0)。
贪心策略:统计牛一共被分成了连续的几段(设有K个位置gap不为0,则牛被分成了连续的K+1段)。然后比较能提供的最大木板数m与K+1的大小:如果m>=K+1,则说明不用覆盖空缺位置就可以完全覆盖牛,此时长度也最短(没有浪费),长度就是牛的个数。如果m<K+1,那么木板不足以把牛每段都独立覆盖,只能把几段合而为一来覆盖,此时就要注意,只需要在覆盖牛的同时覆盖前K+1-m段最短的空缺,就可以满足浪费最少,并且牛被完全覆盖。used数组表示已经被覆盖的空缺段,索引定义与gap相同,为1表示已经覆盖。
代码如下:
/* ID: michael139 LANG: C PROG: barn1 */ #include<stdio.h> #include<string.h> int vis[205],gap[205],used[205]; int main () { FILE *fin = fopen("barn1.in", "r"); FILE *fout = fopen("barn1.out", "w"); int m,s,c,i,j,flag,temp,left,gap_num,diff,ans,min,x; while (fscanf(fin,"%d%d%d",&m,&s,&c) != EOF) { memset(vis,0,sizeof(vis)); memset(gap,0,sizeof(gap)); memset(used,0,sizeof(used)); ans = gap_num = 0; for (i=1;i<=c;i++) { fscanf(fin,"%d",&temp); vis[temp] = 1; } flag = vis[1]; left = 1; for (i=1;i<=s;i++) { if (flag && !vis[i]) { flag = 0; left = i; } else if (!flag && vis[i]) { gap[left-1] = i-left; flag = 1; } } for (i=1;i<=s;i++) { if (gap[i]) { gap_num++; } } diff = gap_num+1-m; if (diff<=0) {//We can get enough boards without cover any empty stalls. for (i=1;i<=s;i++) { if (vis[i]) ans++; } fprintf(fout,"%d\n",ans); } else {//We must choose diff shortest gaps to be covered. while (diff>0) { min = 1000000; for (i=1;i<=s;i++) { if (gap[i] && !used[i] && gap[i]<min) { min = gap[i]; x = i; } } used[x] = 1; for (i=x+1;i<=x+gap[x];i++) vis[i] = 1; diff--; } for (i=1;i<=s;i++) { if (vis[i]) ans++; } fprintf(fout,"%d\n",ans); } } return 0; }