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341 Non-Stop Travel (Dijsktra)

2013-03-08 20:11 323 查看
这是一道很典型的单源点最短路径问题,用Dij来解,记录路径,通过dfs来输出路径

代码如下:

#include <cstdio>
#include <cstring>

const int N = 20;
const int INF = 100000000;
int n, s, t, ans, icase;
int g

, d
, p
;

void dij() {
int mi, v;
for ( int i = 1; i <= n; ++i ) d[i] = INF;
d[s] = 0, ans = 0;
bool vis
;
memset( vis, 0, sizeof(vis) );
for ( int u = 0; u < n; ++u ) {
mi = INF;
for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] < mi ) mi = d[i], v = i;
vis[v] = true;
for ( int i = 1; i <= n; i++ ) if ( !vis[i] && g[v][i] >= 0 && d[i] > d[v] + g[v][i] ) d[i] = d[v] + g[v][i], p[i] = v;
}
}
void output( int x ) {
if ( x == -1 ) return;
output( p[x] );
if ( p[x] != -1 ) printf(" ");
printf("%d", x);
return;
}
int main()
{
icase = 1;
while ( scanf("%d", &n) != EOF && n ) {
for ( int i = 1; i <= n; p[i] = -1, ++i ) for ( int j = i; j <= n; ++j ) g[i][j] = g[j][i] = -1;
for ( int i = 1; i <= n; ++i ) {
int num;
scanf("%d", &num);
for ( int j = 1; j <= num; ++j ) {
int v, del;
scanf("%d %d", &v, &del);
g[i][v] = del;
}
}
scanf("%d%d", &s, &t);
dij();
printf("Case %d: Path = ", icase++);
output( t );
printf("; %d second delay\n", d[t]);
}
}
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