USACO 4.2 Cowcycles(DFS)
2013-03-19 22:13
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使劲水啊。。。数据量太唬人了。。普通DFS就能过。。。看懂题意就好。。看不懂题是硬伤啊。。
/* ID: cuizhe LANG: C++ TASK: cowcycle */ #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <ctime> #include <queue> #include <vector> #include <algorithm> using namespace std; int o1[101],o2[101],ax[15],ay[15]; double arry[101],d[101]; int x,y,x2,y2,n,m; double minz = 100000000; void dfs2(int str,int step) { int i,j,num = 0; double ans = 0,sum = 0; if(step > m) { if(o1 *o2[m] >= 3*o1[1]*o2[1]) { for(i = 1;i <= n;i ++) { for(j = 1;j <= m;j ++) { arry[num++] = (o1[i]*1.0/o2[j]); } } sort(arry,arry+num); for(i = 0;i < num-1;i ++) { d[i] = arry[i+1] - arry[i]; sum += d[i]; } sum = sum/(num-1); for(i = 0;i < num-1;i ++) { ans += (d[i]-sum)*(d[i]-sum); } ans = ans/(num-1); if(minz > ans) { minz = ans; for(i = 1;i <= n;i ++) ax[i] = o1[i]; for(i = 1;i <= m;i ++) ay[i] = o2[i]; } } return ; } for(i = str;i <= y2;i ++) { o2[step] = i; dfs2(i+1,step+1); } } void dfs1(int str,int step) { int i; if(step > n) { dfs2(x2,1); return; } for(i = str;i <= y;i ++) { o1[step] = i; dfs1(i+1,step+1); } return ; } int main() { int i; freopen("cowcycle.in","r",stdin); freopen("cowcycle.out","w",stdout); scanf("%d%d",&n,&m); scanf("%d%d%d%d",&x,&y,&x2,&y2); dfs1(x,1); for(i = 1;i <= n;i ++) { if(i == 1) printf("%d",ax[i]); else printf(" %d",ax[i]); } printf("\n"); for(i = 1;i <= m;i ++) { if(i == 1) printf("%d",ay[i]); else printf(" %d",ay[i]); } printf("\n"); return 0; }
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