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

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

(t��ng) (t��ng) (t��ng) (t��ng) (t��ng) (t��ng) (t��ng) 做ACM 已经有快一�q�了(ji��n)�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�?j��)从一个新生变成了(ji��n)老生�Q�从一个新手变成了(ji��n)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 (t��ng) (t��ng) Azg (t��ng) Taney (t��ng) Stone CB (t��ng) ,他们�l�了(ji��n)我帮助与鼓励�Q�也要感谢我的队友Lvyun Hailer�Q�他们给�?ji��n)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(b��)吧！

Barracuda 2007-03-17 17:34 发表评论