URAL 1957 Wrong Answer 暴力

Wrong Answer

思路：

1.先枚举4的全排列，即球赛的所有可能结果，一共4!=24种情况

2.对于每种情况，DFS 未确定的比赛 的结果，判断这种情况是否可达。

剪枝：

1.对于每种全排列，只需要找到一种满足这个全排列的情况即可，如果有一种情况满足，则不必再枚举其他可能。

2.因为在每种情况下，球队之间的排名已经确定了，所以在DFS的过程中，凡是不满足此排名的情况统统可以剪掉。

这样就可以少很多情况，即使在n=0的情况下也跑出了不错的效率，但是它Wrong Answer on test 17了。。。

求指点T^T

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int MAXN = 10010;
const int permuta[30] = { 1234, 1243, 1324, 1342, 1423, 1432,
2134, 2143, 2314, 2341, 2413, 2431,
3124, 3142, 3214, 3241, 3412, 3421,
4123, 4132, 4213, 4231, 4312, 4321
};

struct Team
{
int id;
int qiu;
int point;
};

bool vis[30];
int game[5][5], gori[5][5];   // 0=i输j 1=i平j 3=i赢j
Team tm[5], ori[5];
int N, cnt;

void show()
{
for ( int i = 1; i <= 4; ++i )
printf( "id=%d point=%d qiu=%d\n", i, tm[i].point, tm[i].qiu );
puts("");
return;
}

void chuli( int i, int j, int c )   //c为队伍i的净赢球量
{
tm[i].point += game[i][j];
tm[j].point += game[j][i];
tm[i].qiu += c;
tm[j].qiu -= c;
return;
}

void init()
{
memset( vis, false, sizeof(vis) );
memset( game, -1, sizeof( game ) );
memset( tm, 0, sizeof(tm) );
for ( int i = 0; i < N; ++i )
{
int a, b, c, d;
scanf( "%d%d%d%d", &a, &b, &c, &d );
if ( c == d )
{
game[a][b] = game[b][a] = 1;
}
else if ( c > d )
{
game[a][b] = 3;
game[b][a] = 0;
}
else
{
game[a][b] = 0;
game[b][a] = 3;
}
chuli( a, b, c - d );
}
return;
}

bool DFS( int *tmp, int cur )
{
if ( cur <= 0 )
{
/*
for ( int i = 1; i <= 4; ++i )
printf( "id=%d point=%d qiu=%d\n", tmp[i], tm[ tmp[i] ].point, tm[ tmp[i] ].qiu );
puts("");
*/
for ( int i = 1; i < 4; ++i )
{
int u = tmp[i];
int v = tmp[i + 1];
if ( tm[u].point < tm[v].point || ( tm[u].point == tm[v].point && tm[u].qiu < tm[v].qiu ) )
return false;
}
return true;
}

int j = tmp[cur];
bool ok = true;
for ( int i = 1; i <= 4; ++i )
{
if ( i == j ) continue;
if ( game[j][i] == -1 )
{
ok = false;
for ( int k = -10; k <= 10; ++k )   //枚举净赢球数
{
if ( k < 0 )
{
game[j][i] = 0;
game[i][j] = 3;
}
else if ( k == 0 )
{
game[j][i] = 1;
game[i][j] = 1;
}
else
{
game[j][i] = 3;
game[i][j] = 0;
}
chuli( j, i, k );

//puts("*************");
//show();

if ( DFS( tmp, cur ) ) return true;

tm[j].point -= game[j][i];
tm[i].point -= game[i][j];
game[j][i] = -1;
game[i][j] = -1;
tm[i].qiu += k;
tm[j].qiu -= k;

//show();
//puts("==============");

}
}
}

if ( ok )
{
if ( cur != 4 )
{
int u = tmp[cur];
int v = tmp[cur + 1];
//printf("%d %d %d %d\n",tm[u].point, tm[v].point, tm[u].qiu, tm[v].qiu );
if ( tm[u].point < tm[v].point || ( tm[u].point == tm[v].point && tm[u].qiu < tm[v].qiu ) )
return false;
else
{
if ( DFS( tmp, cur - 1 ) ) return true;
}
}
else
{
if ( DFS( tmp, cur - 1 ) ) return true;
}
}
return false;
}

void solved()
{
for ( int i = 1; i <= 4; ++i )
{
ori[i] = tm[i];
for ( int j = 1; j <= 4; ++j )
gori[i][j] = game[i][j];
}

int temp[5];
for ( int i = 0; i < 24; ++i )
{
int nn = permuta[i];
for ( int j = 4; j > 0; --j )
{
temp[j] = nn % 10;
nn /= 10;
}
/*
for ( int j = 1; j <= 4; ++j )
printf( "%d ", temp[j] );
puts("");
*/
for ( int j = 1; j <= 4; ++j )
{
tm[j] = ori[j];
for ( int k = 1; k <= 4; ++k )
game[j][k] = gori[j][k];
}

if ( DFS( temp, 4 ) )
{
++cnt;
vis[i] = true;
}

//      puts("-------");
}
}

int main()
{
//freopen("s.out", "w", stdout );
while ( ~scanf( "%d", &N ) )
{
init();

cnt = 0;
solved();

int ans[4];
printf( "%d\n", cnt );
for ( int i = 0; i < 24; ++i )
if ( vis[i] )
{
int nn = permuta[i];
for ( int j = 3; j >= 0; --j )
{
ans[j] = nn % 10;
nn /= 10;
}
for ( int j = 0; j < 4; ++j )
{
if ( j ) putchar(' ');
printf( "%d", ans[j] );
}
puts("");
}
}
return 0;
}