Massive Algorithms: POJ 3169 Layout

http://poj.org/problem?id=3169

Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).

Some cows like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows dislike each other and the minimum distance by which they must be separated.

Your job is to compute, if possible, the maximum possible distance between cow 1 and cow N that satisfies the distance constraints.

Input

Line 1: Three space-separated integers: N, ML, and MD.

Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.

Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.

Output

Line 1: A single integer. If no line-up is possible, output -1. If cows 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between cows 1 and N.

Sample Input

Sample Output

Hint

Explanation of the sample:

There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.

The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.

http://blog.csdn.net/u012860063/article/details/46862131

http://www.hardbird.net/poj-3169-layout%E5%B7%AE%E5%88%86%E7%BA%A6%E6%9D%9F/

大意：n只牛排成一列，编号1到n。某些牛之间有两中关系：两只牛i和j相互喜欢，之间的距离不超过d；两只牛i和j相互厌恶，之间的距离不能小于d。问在符合所以关系的情况下，牛n距离牛1最远多远。

思路：明显的差分约束问题，对于两种关系分别建边即可。

喜欢关系：i<j: d[j]-d[i]<=d, 建边i-j，权值为d.

厌恶关系：i<j: d[i]-d[j]<=-d, 建边j-i, 权值为-d.

令d[1]=0, 如果存在负环，就是不满足所有关系，输出-1；如果d[n]的取值不受任何关系的约束，输出-2；如果存在可行解，答案就是d[n].

#define INF 0x3f3f3f3f
#define N 20000
#define M 20000
int n, m, k;
int Edgehead[N], dis[N];
struct Edge
{
    int v,w,next;
} Edge[2*M];
bool vis[N];
int cont[N];
void Addedge(int u, int v, int w)
{
    Edge[k].next = Edgehead[u];
    Edge[k].w = w;
    Edge[k].v = v;
    Edgehead[u] = k++;
}
int SPFA( int start)//stack
{
    int sta[N];
    memset(cont,0,sizeof(cont);
    int top = 0;
    for(int i = 1 ; i <= n ; i++ )
        dis[i] = INF;
    dis[start] = 0;
    ++cont[start];
    memset(vis,false,sizeof(vis));
    sta[++top] = start;
    vis[start] = true;
    while(top)
    {
        int u = sta[top--];
        vis[u] = false;
        for(int i = Edgehead[u]; i != -1; i = Edge[i].next)//注意
        {
            int v = Edge[i].v;
            int w = Edge[i].w;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u]+w;
                if( !vis[v] )//防止出现环
                {
                    sta[++top] = v;
                    vis[v] = true;
                }
                if(++cont[v] > n)//有负环
                    return -1;
            }
        }
    }
    return dis[n];
}
int main()
{
    int u, v, w;
    int c;
    int ml, md;
    while(~scanf("%d%d%d",&n,&ml,&md))//n为目的地
    {
        k = 1;
        memset(Edgehead,-1,sizeof(Edgehead));
        for(int i = 1 ; i <= ml; i++ )
        {
            scanf("%d%d%d",&u,&v,&w);
            Addedge(u,v,w);
        }
        for(int i = 1 ; i <= md; i++ )
        {
            scanf("%d%d%d",&u,&v,&w);
            Addedge(v,u,-w);
        }
        for(int i = 1; i < n; i++)
        {
            Addedge(i+1,i,0);
        }
        int ans = SPFA(1);//从点1开始寻找最短路
        if(ans == INF)
        {
            printf("-2\n");
        }
        else
        {
            printf("%d\n",ans);
        }
    }
    return 0;
}

https://gist.github.com/KuoE0/1950504
struct EDGE {
int v, w, next;
};

EDGE edge[ 3000010 ];
int adj[ MAXN ], dis[ MAXN ], cnt[ MAXN ];
bool inque[ MAXN ];
int n, e1, e2, total;
queue< int > que;

void addEdge( int a, int b, int w ) {
edge[ total ].v = b, edge[ total ].w = w, edge[ total ].next = adj[ a ];
adj[ a ] = total;
++total;
}

int SPFA( int sr ) {
int now, next, i;
memset( dis, 63, sizeof( dis ) );
memset( cnt, 0, sizeof( cnt ) );
while ( !que.empty() )
que.pop();
dis[ sr ] = 0;
inque[ sr ] = 1, cnt[ sr ] = n;
que.push( sr );

while ( !que.empty() ) {
now = que.front();
que.pop();
inque[ now ] = 0;
for ( i = adj[ now ]; i != -1; i = edge[ i ].next ) {
EDGE temp = edge[ i ];
next = edge[ i ].v;
if ( dis[ now ] + edge[ i ].w < dis[ next ] ) {
dis[ next ] = dis[ now ] + edge[ i ].w;
if ( !inque[ next ] ) {
++cnt[ next ];
if ( cnt[ next ] > n )
return -1;

que.push( next );
inque[ next ] = 1;
}
}
}
}
return dis[ n ] != dis[ 0 ] ? dis[ n ] : -2;
}

int main() {
int i, a, b, w;

while ( ~scanf( "%d %d %d", &n, &e1, &e2 ) ) {
total = 0;
memset( adj, -1, sizeof( adj ) );
for ( i = 0; i < e1; ++i ) {
scanf( "%d %d %d", &a, &b, &w );
addEdge( a, b, w );
}
for ( i = 0; i < e2; ++i ) {
scanf( "%d %d %d", &a, &b, &w );
addEdge( b, a, -w );
}
printf( "%d\n", SPFA( 1 ) );
}
return 0;
}

POJ 3169 Layout

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