|
#
stanford open course： machine learning
實變函數
測度論與概率論
隨機過程
算法導論
新概念英語
昨晚發過去的，今天一早就給出了反饋信息?？赡苁侵皫熜肿龅貌畈欢嗔?，所以只有兩點針對證明的意見，下面是 Wei_Fan的回信： Some clarifications need to be made to the formal analyses: 1. It is not clearly to me, in Theorem 1, what exactly is b. This needs to be defined formally and clearly. I think that this is the bayesian optimal decision. 2. I do not understand why conf(x) = max p(y|x)? In reality, isn't the confidence of a prediction the estimated probability by a model M, and there is always a dependency on M? In other words, the estimated probability by a model is p(y|x,M) and there is an explicit dependency on M, and normally P(y|x) = P(y|x,M). I think that there is one more step need to be done. That is to assume M is better than random guessing, thus, P(y|x,M) is reasonably close to P(y|x),..
摘要: Dijkstra with Heap 閱讀全文
The Perfect Stall
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 2136 |
|
Accepted: 961 |
Description
Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall.
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.
Input
The input includes several cases. For each case, the first line contains two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. Each of the following N lines corresponds to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow.
Output
For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made.
Sample Input
5 5
2 2 5
3 2 3 4
2 1 5
3 1 2 5
1 2
Sample Output
4
 #include <iostream>
using namespace std;
int map[201][201];
int match[201];
int visit[201];
int n,m;
bool dfs(int p)
   {
 int x;
for(x=0;x<m;x++)
  {
if(map[p][x]==1 && visit[x]==-1)
{
visit[x]=1;
int t=match[x];
match[x]=p;
if(t==-1 || dfs(t)) return true;
match[x]=t;
 }
}
return false;
 }
int find_match()
{
memset(match,-1,sizeof(match));
int i,sum=0;
for(i=0;i<n;i++)
{
memset(visit,-1,sizeof(visit));
if(dfs(i)) sum++;
}
return sum;
}
int main()
{
//freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&m)!=EOF)
{
int i,j;
for(i=0;i<n;i++)
for(j=0;j<m;j++) map[i][j]=0;
for(i=0;i<n;i++)
{
int a,b;
scanf("%d",&a);
for(j=0;j<a;j++)
{
scanf("%d",&b);
map[i][b-1]=1;
}
}
int res=find_match();
printf("%d\n",res);
}
return 1;
}
參考了郭嵩山的第三冊，用網上的代碼似乎有錯。
記錄一下2007-12-22的事： 昨天開始跟 Wei Fan進行討論，一個IBM Research的牛人，主要是說出我們在各自的領域里面的想法，以及討論寫論文的東西。 昨天施瀟瀟師兄講了domain transfer的一些情況，接著他講了他的幾個想法，主要還是兩個領域個性與共性的問題。 另外，Fan老師的一些建議：考慮out-of-domain的active learning；想法必須簡單，intuitive，在五分鐘內能跟別人說清楚；想大的topic；考慮將graph spectra用在transfer learning上；提的問題要令人exciting，起點要高；優化算法要懂一些；看paper時不要陷入到細節中，要從高的角度來看文章。 最后跟他用英文聊了一會，發現我英文真是爛啊。
poj兩個相似的題目，很有代表性。
pku 1828
pku 2726
題目大意是給出平面上的若干點，找出符合某種要求的點(x0, y0)，如要求不存在點(x, y)，使得x>=x0 且 y>=y0，換一種說法就是對任意點(x, y)都有x<x0 或 y<y0. 下面給出pku 1828的代碼
#include <stdlib.h>
#include <stdio.h>
struct node{
int x,y;
}a[50001];
int cmp(const void *aa, const void *bb)
{
node* a=(node*)aa;
node* b=(node*)bb;
if(a->x==b->x) return a->y-b->y;
return a->x-b->x;
}
int i,j,k,n;
int main()
{
while(scanf("%d",&n) && n!=0)
{
for(i=0;i<n;i++) scanf("%d%d",&a[i].x, &a[i].y);
qsort(a,n,sizeof(node),cmp);
//for(i=0;i<n;i++) printf("%d %d\n",a[i].x,a[i].y);
int total=1;
int maxi=a[n-1].y;
for(i=n-2;i>=0;i--)
{
if(a[i].y>maxi)
{
maxi=a[i].y;
total++;
}
}
printf("%d\n",total);
}
return 1;
}
|