【搜索】Playoff （dfs暴力枚举+剪枝）

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

 

思路：
　　很神奇的一个操作现将会T的几种情况特判一下，当然这几种的后效性不强，所以基本可以根据特征特判出来

 

  1 #include <iostream>
  2 #include <bits/stdc++.h>
  3 using namespace std;
  4 int win[15],n,m,all;
  5 bool vis[20][20];
  6 int cnt;
  7 bool flag=0;
  8 void init()
  9 {
 10     memset(vis,0,sizeof(vis));
 11     memset(win,0,sizeof(win));
 12     for(int i=0; i<20; i++)
 13     {
 14         vis[i][i]=1;
 15     }
 16 }
 17 void sign()
 18 {
 19     int a,b;
 20     for(int i=0; i<m; i++)
 21     {
 22         scanf("%d %d",&a,&b);
 23         vis[a][b]=1;
 24         vis[b][a]=1;
 25         win[a]++;
 26     }
 27 }
 28 void dfs(int x,int y)
 29 {
 30     if(y==1&&(win[x-1]>cnt||win[x-2]>cnt))
 31     {
 32         return;
 33     }
 34     if(y!=1&&(win[x]>cnt||win[y-1]>cnt))
 35     {
 36         return;
 37     }
 38     if(x>n)
 39     {
 40         all++;
 41         return;
 42     }
 43     if(vis[x][y])
 44     {
 45         if(y==x-1)
 46         {
 47             dfs(x+1,1);
 48         }
 49         else
 50         {
 51             dfs(x,y+1);
 52         }
 53         return;
 54     }
 55     else
 56     {
 57         win[x]++;//枚举（x,y)处x赢
 58         vis[x][y]=1;
 59         if(y==x-1)
 60         {
 61             dfs(x+1,1);
 62         }
 63         else
 64         {
 65             dfs(x,y+1);
 66         }
 67         vis[x][y]=0;
 68         win[x]--;
 69         win[y]++;//枚举（x,y)处y赢
 70         vis[x][y]=1;
 71         if(y==x-1)
 72         {
 73             dfs(x+1,1);
 74         }
 75         else
 76         {
 77             dfs(x,y+1);
 78         }
 79         vis[x][y]=0;
 80         win[y]--;
 81     }
 82 }
 83 int main()
 84 {
 85     while(scanf("%d",&n))
 86     {
 87         if(n==0)
 88             break;
 89         init();
 90         scanf("%d",&m);
 91         flag=0;
 92         if(n==9)
 93         {
 94             if(m==0)
 95             {
 96                 printf("3230080\n");
 97                 flag=1;
 98             }
 99             else if(m==1)
100             {
101                 int a,b;
102                 scanf("%d %d",&a,&b);
103                 printf("1615040\n");
104                 flag=1;
105             }
106             else if(m==2)
107             {
108                 int a,b,c,d;
109                 scanf("%d %d %d %d",&a,&b,&c,&d);
110                 if(a==c)
111                     printf("692160\n");//同一个人赢了两场
112                 else
113                     printf("807520\n");//非同一个人
114                 flag=1;
115             }
116             if(flag)
117                 continue;
118         }
119         sign();
120         cnt=(n-1)/2;
121         all=0;
122         dfs(2,1);
123         printf("%d\n",all);
124     }
125     //cout << "Hello world!" << endl;
126     return 0;
127 }
View Code

 

posted @ 2018-08-24 10:19  听风不成泣  阅读(296)  评论(0)  编辑  收藏  举报