题目描述
The Minato Mirai Football Association hosts its annual championship as a single round-robin tournament, in which each team plays a single match against all the others. Unlike many other round-robin tournaments of football, matches never result in a draw in this tournament. When the regular time match is a tie, overtime is played, and, when it is a tie again, a penalty shootout is played to decide the winner.
If two or more teams won the most number of matches in the round-robin, a playoff is conducted among them to decide the champion. However, if the number of teams is an odd number, it is possible that all the teams may have the same number of wins and losses, in which case all the teams participate in the playoff, called a “full playoff” here.
Now, some of the tournament matches have already been played and we know their results. Whether or not a full playoff will be required may depend on the results of the remaining matches. Write a program that computes the number of win/loss combination patterns of the remaining matches that lead to a full playoff.
The first datatset of the Sample Input represents the results of the first three matches in a round-robin tournament of five teams, shown in the following table. In the table, gray cells indicate the matches not played yet.
In this case, all the teams win the same number of matches with only two win/loss combination patterns of the remaining matches, which lead to a full playoff, as shown below. In the two tables, the differences are indicated in light yellow.
输入
The input consists of multiple datasets, each in the following format.
n
m
x1 y1
…
xm ym
n is an odd integer, 3, 5, 7, or 9, indicating the number of teams participating in the tournament. m is a positive integer less than n(n−1)/2, which is the number of matches already finished. xi and yi give the result of the i-th match that has already taken place, indicating that team xi defeated team yi. Each of xi and yi is an integer 1 through n which indicates the team number. No team plays against itself, that is, for any i, xi ≠ yi. The match result of the same team pair appears at most once. That is, if i ≠ j, then (xi,yi) ≠ (xj,yj) and (xi,yi) ≠ (yj,xj) hold.
The end of the input is indicated by a line containing a zero. The number of datasets does not exceed 100.
输出
For each dataset, output a single line containing one integer which indicates the number of possible future win/loss patterns that a full playoff will be required.
样例输入
5
3
3 2
4 1
5 1
3
1
1 2
3
2
1 2
3 2
5
4
4 1
4 2
5 1
5 2
5
3
4 1
4 2
5 1
5
4
3 2
4 1
5 1
5 2
9
11
6 1
6 4
7 2
7 3
7 4
8 2
8 3
8 4
9 1
9 3
9 5
9
10
6 1
6 4
7 2
7 3
7 4
8 2
8 3
8 4
9 1
9 3
5
6
4 3
2 1
5 1
2 4
1 3
2 3
9
1
1 2
0
样例输出
2
1
0
0
1
0
0
16
0
1615040
题意：有N个队比赛，一个队要和剩下的所有队比一次，现在限定，每个队打的N-1场比赛中必须有一半赢一般输。给出的N都是奇数。给出一些队伍的已知成绩，给剩余空白部分填输赢，问有多少种合法的填法。
一开始以为有什么优秀的类似于组合数的写法，找了半天公式并没有什么收获。
最后没办法的办法，搜吧。其实仔细一想也没什么不能搜的，因为题目数据量只有3，5，7，9四个值。并且一个样例已经给出了9个队时有1615040种，那么9个队一个都不是已知结果时最多是这个样例*2。那么直接搜索剪枝即可，一定要剪枝，尝试了多次后已经能保证搜索最后跑出来的结果必定是一组可行解。对于每个队，只有输和赢两种可能，而对于每个已知部分，不分支即可。要保证符合条件，那么一些给出的结果首先是可以退出另一些结果的，在输入后根据已知结果最大限度推出必定的为止结果，对于剩余空白部分枚举输赢就行了。
#include<bits/stdc++.h>
#define LL long long
#define M(a,b) memset(a,b,sizeof a)
#define pb(x) push_back(x)
using namespace std;
const int maxn=20;
int n,k,ans,n2;
int won[maxn],def[maxn];
int mp[maxn][maxn];
void dfs(int x,int y)
{
for(int i=1; i<=n; i++)if(won[i]>n2||def[i]>n2)return;///剪枝判断，不符合输赢数量条件的直接剪了
if(x>n)
{
// for(int i=1; i<=n; i++)
// for(int j=1; j<=n; j++)
// printf("%d%c",mp[i][j],j==n?'
':' ');
// printf("==============================
");
for(int i=1; i<=n; i++)
if(won[i]!=n2||def[i]!=n2)return;
ans++;
return;
}
if(x==y)return;
if(mp[x][y]==0)
{
won[x]++;
def[y]++;
mp[x][y]=1;
mp[y][x]=-1;
dfs(y==x-1?x+1:x,y==x-1?1:y+1);
mp[x][y]=mp[y][x]=0;
won[x]--;
def[y]--;
def[x]++;
won[y]++;
mp[x][y]=-1;
mp[y][x]=1;
dfs(y==x-1?x+1:x,y==x-1?1:y+1);///相当于用搜索走了一个for循环
mp[x][y]=mp[y][x]=0;///取消标记
def[x]--;
won[y]--;
}
else dfs(y==x-1?x+1:x,y==x-1?1:y+1);
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(!n)return 0;
else scanf("%d",&k);
n2=n>>1;
M(won,0);
M(def,0);
M(mp,0);
ans=0;
bool flag=false;
int x,y;
while(k--)
{
scanf("%d%d",&x,&y);
if(!mp[x][y])
{
mp[x][y]=1;
mp[y][x]=-1;
won[x]++;
def[y]++;
}
if(won[x]>n2||def[y]>n2)flag=true;
}
if(!flag)///处理输入后对已成定局的未知结果的标记
{
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
if(i==j)continue;
if(won[x]==n2)
for(int i=1; i<=n; i++)
if(mp[x][i]==0&&x!=i)
{
mp[x][i]=-1;
mp[i][x]=1;
won[i]++;
def[x]++;
}
if(def[y]==n2)
for(int i=1; i<=n; i++)
if(mp[y][i]==0&&y!=i)
{
mp[y][i]=1;
mp[i][y]=-1;
def[i]++;
won[y]++;
}
}
}
dfs(2,1);
}
printf("%d
",ans);
}
}