很裸的一道最小瓶颈路的模板题,先求最小生成树,然后在求点对点的最小瓶颈路
代码如下:
#include
#include
#include
#include
using namespace std;
const int N = 210;
const double INF = 1000000;
int n, p[N];
double x[N], y[N], map[N][N], f[N][N], s1;
bool mark[N];
void prim() {
bool vis[N];
memset(vis, 0, sizeof(vis));
double d[N], mi;
for ( int i = 1; i <= n; ++i ) d[i] = INF, p[i] = -1;
d[1] = 0.0, s1 = 0;
int v;
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;
s1 += mi;
for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] > map[v][i] ) d[i] = map[v][i], p[i] = v;
}
//printf("%lf\n", s1);
}
void dfs( int v ) {
for ( int u = 1; u <= n; ++u ) {
if ( !mark[u] && p[u] == v ) {
mark[u] = true;
for ( int x = 1; x <= n; ++x )
if ( mark[x] && x != u ) f[x][u] = f[u][x] = max( f[x][v], map[u][v] );
dfs(u);
}
}
}
int main()
{
int icase = 1;
while ( scanf("%d", &n), n) {
for ( int i = 1; i <= n; ++i )
scanf("%lf%lf", &x[i], &y[i]);
for ( int i = 1; i <= n; ++i ) for ( int j = i+1; j <= n; ++j )
map[i][j] = map[j][i] = sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
prim();
memset(mark, 0, sizeof(mark));
for ( int i = 1; i <= n; ++i ) for ( int j = 1; j <= n; ++j ) f[i][j] = 0.0;
mark[1] = true;
dfs( 1 );
//for ( int i = 1; i <= n; printf("\n"), ++i ) for ( int j = 1; j <= n; ++j ) printf("%lf ", f[i][j]);
printf("Scenario #%d\n", icase++);
printf("Frog Distance = %.3lf\n\n", f[1][2]);
}
}