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Barracuda — Wed, 28 Mar 2007 10:52:00 GMT

和自��q��比�v来，感觉大牛的代码要�_�悍的多啊�?br />代码如下�Q?br />#include
#include
#define maxn 250
struct Map
{
int f;
int c;
}map[maxn][maxn];
int pre[maxn];
int q[maxn*maxn];
int v[maxn];
int N,M;
int s,t;
int abs( int x ){ return x > 0 ? x : -x ; }
int min( int x, int y ){ return x < y ? x : y; }
void init()
{
int i, S, E, C;
memset( map, 0, sizeof(map) );
for(i=0;i {
  scanf( "%d%d%d", &S, &E, &C );
  map[S][E].c += C;
}
}
void solve()
{
int i,j;
int head,tail;
s = 1;
t = M;
while(true)
{
  memset( pre, 0, sizeof(pre) );
  head = 0, tail = 1;
  q[0] = s;
  v[s] = 1000000000;
  pre[s] = s;
  while( head < tail && pre[t] == 0 )
  {
   i = q[head];
   for( j = 1; j <= M; j++ )
   {
    if( pre[j] == 0 )
    {
     if( map[i][j].f < map[i][j].c )
      pre[j] = i , q[tail++] = j , v[j] = min( v[i], map[i][j].c-map[i][j].f );
     else if( map[j][i].f > 0 )
      pre[j] = -i, q[tail++] = j , v[j] = min( v[i], map[j][i].f );
    }
            }
   head++;
  }
  if( pre[t] == 0 )break;

  i = t;
  while( i != s )
  {
   j = abs( pre[i] );
   if( pre[i] > 0 )map[j][i].f += v[t];
   else map[i][j].f -= v[t];
   i = j;
  }
}
int ans = 0;
for( i = 1; i <= M; i++ )ans += map[s][i].f;
printf("%d\n",ans);
}
int main()
{
while(scanf("%d%d",&N,&M)!=EOF)
{
  init();
  solve();
}
return 0;
}

Barracuda 2007-03-28 18:52 发表评论

Barracuda — Thu, 22 Mar 2007 07:18:00 GMT

�W�一�ơ写�|�络最大流��法�Q�居然做对了�?br>该算法要用到 Dijkstra ��法�?br>每次用Dijkstra��法求的一条可通过的�\后，扑և�该�\上面的最��的权�?MIN�Q�然后将该�\径上的每条边的权�?减去 MIN �Q�反方向的权值加 MIN �Q?例如�Q�e( v1, v2 ) 是�\径上面的一条边�Q�则Map[v1][v2] -= MIN, Map[v2][v1] += MIN ). 累计MIN�Q?其最后的值就是所要求的最大流�?br>当最�?在图中找不到�q�接 1, n 两点的�\径时�Q�算法完毕�?br>
//2019083 whitesea 1273 Accepted 224K 15MS C++ 1880B 2007-03-22 14:56:03
// PKU 1273 �|�络最大流
http://acm.pku.edu.cn/JudgeOnline/problem?id=1273

#include
#include
#define MAX 210
int Map[MAX][MAX];
int n, M, SUM, Path[MAX], distance[MAX], Min;
int IN(){
    int i, x, y, d;
    if( scanf( "%d %d", &M, &n ) == EOF )return 0; // M �?边的数目�Q?n 是点的数�?br>    memset( Map, 0, sizeof( Map ) );
    for( i = 1; i <= M; i++ ){
        scanf( "%d %d %d", &x, &y, &d );   // 点x 到点y��量�?d
        Map[x][y] += d;
    }
    return 1;
}
void Dijkstra(){
    int mindis, i, j, u, s[MAX];
    for( i = 1; i <= n; i++ ){
        distance[i] = Map[1][i];
        s[i] = 0;

        if( i != 1 && distance[i] > 0 )Path[i] = 1;
        else Path[i] = -1;
    }
    s[1] = 1;
    for( i = 2; i <= n; i++ ){
        mindis = 0;
        for( j = 1; j <= n; j++ ){
            if( s[j] == 0 && distance[j] > mindis ){
                u = j;
                mindis = distance[j];
            }
        }
        if( mindis == 0 )return;
        s[u] = 1;
        for( j = 1; j <= n; j++ ){
            if( s[j] == 0 && Map[u][j] > 0 && distance[u] + Map[u][j] > distance[j] ){
                distance[j] = distance[u] + Map[u][j];
                Path[j] = u;
            }
        }
    }
}

void Find(){
    int s, t;
    Min = 2000000000;
    s = n;
    t = Path[n];
    while( t != -1 ){
        if( Map[t][s] < Min )Min = Map[t][s];
        s = t;
        t = Path[t];
    }
}

void Change(){
    int s, t;
    s = n;
    t = Path[n];
    while( t != -1 ){
        Map[t][s] -= Min;
        Map[s][t] += Min;
        s = t;
        t = Path[t];
    }
}

void SOLVE(){
    int i, j, k;
    SUM = 0;
    while( 1 ){
        Dijkstra();
        if( Path[n] == -1 )return;
        Find();
        SUM += Min;
        Change();
    }
}

void OUT(){
printf( "%d\n", SUM );
}

int main(){
    while( IN() ){

        SOLVE();
        OUT();
    }
    return 0;
}

Barracuda 2007-03-22 15:18 发表评论

做PKU做ACM �?00题纪念帖

Barracuda — Sat, 17 Mar 2007 09:34:00 GMT

做ACM 已经有快一�q�了�Q�终于在PKU做满200题，发帖�U�念下，鼓励自己再接再厉。这一�q�发生的变化�q�真大啊�Q�一不小心从一个新生变成了老生�Q�从一个新手变成了老队员。想��h��q�一�q�来�Q�感觉收莯��是很大的。在�q�里我要特别感谢XC Azg Taney Stone CB ,他们�l�了我帮助与鼓励�Q�也要感谢我的队友Lvyun Hailer�Q�他们给了我很多的帮助。在做ACM �q�个斚w��Q�我们学校一直很差，�?6~07的全国三个赛区都��|而归。现在我得努力，我的队友得努力，CUG 的ACMER 都得努力。让我们来把CUG 的ACM 做好�Q?努力吧，加��a吧！

Barracuda 2007-03-17 17:34 发表评论