## Sunday, January 10, 2016

### 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
```4 2 1
1 3 10
2 4 20
2 3 3```
Sample Output
`27`
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.

```#define INF 0x3f3f3f3f
#define N 20000
#define M 20000
int n, m, k;
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].w = w;
Edge[k].v = v;
}
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;
for(int i = 1 ; i <= ml; i++ )
{
scanf("%d%d%d",&u,&v,&w);
}
for(int i = 1 ; i <= md; i++ )
{
scanf("%d%d%d",&u,&v,&w);
}
for(int i = 1; i < n; i++)
{
}
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 ];
++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;
for ( i = 0; i < e1; ++i ) {
scanf( "%d %d %d", &a, &b, &w );
}
for ( i = 0; i < e2; ++i ) {
scanf( "%d %d %d", &a, &b, &w );
}
printf( "%d\n", SPFA( 1 ) );
}
return 0;
}