这道题,wa了,原因是没有需要建路的时候,输出一行提示,我忽视了
水题,唯一有点特别的是需要输出新建的边,那么就要建一棵前驱树,用数组表示
代码:
#include <cstdio> #include <cstring> #include <cmath> const int N = 800; const double INF = 10000000000; int T, n, m, num, p[N]; double map[N][N], d[N], x[N], y[N]; struct highway{ int u, v; }way[N]; void prim() { for ( int i = 0; i <= n; ++i ) d[i] = INF, p[i] = -1; double mi; d[1] = 0; bool vis[N]; int v; 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 ) { v = i, mi = d[i]; } vis[v] = true; if ( mi > 0 ) { printf("%d %d\n", v, p[v]); num++; } for ( int i = 1; i <= n; ++i ) if ( !vis[i] && d[i] - map[v][i] > 0.0 ) { d[i] = map[v][i]; p[i] = v; //printf("%d %d\n", i, p[i] ); } } } int main() { scanf("%d", &T); while ( T-- ) { scanf("%d", &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 ){ double tmp = (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]); map[j][i] = map[i][j] = tmp; } scanf("%d", &m); while ( m-- ) { int s,t; scanf("%d%d", &s, &t); map[s][t] = map[t][s] = 0; } num = 0; prim(); if ( num == 0 ) printf("No new highways need\n"); if ( T > 0 ) printf("\n"); } }