POJ 2484(博弈-对称博弈)

A Funny Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3345   Accepted: 1960

Description

Alice and Bob decide to play a funny game. At the beginning of the game they pick n(1 <= n <= 106) coins in a circle, as Figure 1 shows. A move consists in removing one or two adjacent coins, leaving all other coins untouched. At least one coin must
be removed. Players alternate moves with Alice starting. The player that removes the last coin wins. (The last player to move wins. If you can't move, you lose.) 
 

Note: For n > 3, we use c1, c2, ..., cn to denote the coins clockwise and if Alice remove c2, then c1 and c3 are NOT adjacent! (Because there is an empty place between c1 and c3.) 

Suppose that both Alice and Bob do their best in the game. 
You are to write a program to determine who will finally win the game.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (1 <= n <= 106). There are no blank lines between cases. A line with a single 0 terminates the input. 

Output

For each test case, if Alice win the game,output "Alice", otherwise output "Bob". 

Sample Input

1
2
3
0

Sample Output

Alice
Alice
Bob

Source

POJ Contest,Author:Mathematica@ZSU

博弈第一招——对称博弈。

思路:无论对方怎么取,我都取与它对称(或者相反/一样的)

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
	int n;
	while(scanf("%d",&n)&&n)
	{
		if (n>=3) cout<<"Bob"<<endl;
		else cout<<"Alice"<<endl;
	}
	return 0;
}

POJ 2362(Square-搜索剪枝1-相对顺序)

Language:
Square
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 17066   Accepted: 5878

Description

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

Input

The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

Output

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

Sample Input

3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5

Sample Output

yes
no
yes

Source

《搜索是怎样剪枝的-1》

1.只要找到3条边。

2.从大到小顺序找。

3.每次搜边时要确定边的相对顺序唯一(直接从TLE→秒)


#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cstring>
#include<iostream>
using namespace std;
#define MAXM (20+10)
int tt,n,a[MAXM],cnt,len;
bool b[MAXM],flag;
bool dfs(int l,int x,int pre)
{
//	cout<<l<<' '<<x<<' '<<kth<<endl;
	if (x==len) {l++;x=0;pre=n-1;}
	if (l==4)
	{
		return 1;
	}
	for(int i=pre-1;i;i--)
		if (!b[i]&&x+a[i]<=len)
		{
			b[i]=1;
			if (dfs(l,x+a[i],i)) return 1;
			b[i]=0;
		//	if (!x) return 0;
		}
	return 0;
}
int main()
{
	scanf("%d",&tt);
	while (tt--)
	{
		cnt=0;
		scanf("%d",&n);
		for (int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			cnt+=a[i];b[i]=0;
		}
		sort(a+1,a+1+n);
		if (n<4||cnt%4||a[n]>cnt/4)
		{
			printf("non");continue;
		}
		b[n]=1;len=cnt/4;
		if (dfs(1,a[n],n))
		{
			printf("yesn");
		}
		else printf("non");
	}
	return 0;
}


POJ 2553(The Bottom of a Graph-缩点求出度)

Language:
The Bottom of a Graph
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 7308   Accepted: 3003

Description

We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges.
Then G=(V,E) is called a directed graph. 
Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1).
Then p is called a path from vertex v1 to vertex vn+1 in G and we say that vn+1 is reachable from v1, writing (v1→vn+1)
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from vv is also reachable from w. The bottom of a graph is the subset of
all nodes that are sinks, i.e.,bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.

Input

The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer
numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v1,w1,...,ve,we with
the meaning that (vi,wi)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

Output

For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.

Sample Input

3 3
1 3 2 3 3 1
2 1
1 2
0

Sample Output

1 3
2

Source

Tarjen缩点统计出度。


#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cstring>
#include<iostream>
using namespace std;
#define MAXN (5000+10)
#define MAXM (1000000+10)
int n,m,h[MAXN],t[MAXN],dfs[MAXN],tim,kind,s[MAXN],tot;
bool b[MAXN];
int edge[MAXM],next[MAXM],pre[MAXN],size;
int x[MAXM],y[MAXM],numk[MAXN];
void addedge(int u,int v)
{
	edge[++size]=v;
	next[size]=pre[u];
	pre[u]=size;
}
void tar(int u)
{
	dfs[u]=t[u]=++tim;b[u]=1;s[++tot]=u;
	for (int p=pre[u];p;p=next[p])
	{
		int &v=edge[p];
		if (!b[v]) tar(v),dfs[u]=min(dfs[u],dfs[v]);
		else if (!h[v]) dfs[u]=min(dfs[u],t[v]);  //保证指向祖先
	}
	if (dfs[u]==t[u])
	{
		kind++;
		while (s[tot]!=u) h[s[tot--]]=kind,numk[kind]++;
		h[s[tot--]]=kind;numk[kind]++;

	}
}
int outdegree[MAXN];
int main()
{
	while (scanf("%d",&n)&&n)
	{
		scanf("%d",&m);
		tot=size=tim=kind=0;
		memset(h,0,sizeof(h));
		memset(t,0,sizeof(t));
		memset(dfs,0,sizeof(dfs));
		memset(s,0,sizeof(s));
		memset(pre,0,sizeof(pre));
		memset(b,0,sizeof(b));
		memset(numk,0,sizeof(numk));
		memset(outdegree,0,sizeof(outdegree));

		for (int i=1;i<=m;i++)
		{
			scanf("%d%d",&x[i],&y[i]);
			addedge(x[i],y[i]);
		}
		for (int i=1;i<=n;i++) if (!b[i]) tar(i);
//		for (int i=1;i<=n;i++) cout<<h[i]<<' ';cout<<endl;
		for (int i=1;i<=m;i++)
		{
			if (h[x[i]]^h[y[i]]) outdegree[h[x[i]]]++;
		}
		/*
		int ans=0;
		for (int i=1;i<=kind;i++)
			if (!outdegree[i]) ans+=numk[i];
		cout<<ans<<endl;
		*/
		bool flag=0;
		for (int i=1;i<=n;i++)
			if (!outdegree[h[i]])
			{
				if (flag) printf(" ");else flag=1;printf("%d",i);
			}


	//	cout<<ans<<endl;
		cout<<endl;
	}
	return 0;
}