PKU POJ 1679

在prim过程中，判断是否存在产生最小生成树的不同途径，分两种情况：
1.取最小d[i]时,判断是否存在d[j]==d[i](i!=j);如果d[j]再之后改变了,说明在之前走i和j会生成两棵不同的MST;
2.更新d[i]=D时,判断从已确定节点到i的距离中有k个为d[i],k>1时,则当i以D的距离被添加到生成树中时,说明这k条边可任意替换,Not Unique!

#include<iostream> using namespace std; #define S 101 int map[S][S], d[S]; bool vis[S], mark[S], v[S]; int N, M, res; bool prim() { memset( vis, 0, sizeof vis ); memset( mark, 0, sizeof mark ); memset( v, 0, sizeof v ); int i, j, pos; res = 0; for( i = 2; i <= N; ++i ) d[i] = map[1][i]; d[1] = 0; vis[1] = true; for( i = 0; i <= N; ++i ){ pos = 0; for( j = 2; j <= N; ++j ){ if( vis[j] || d[j] == -1 ) continue; if( pos == 0 || d[pos] > d[j] ) pos = j; } if( pos == 0 ) return true; if( v[pos] ) return false; vis[pos] = true; for( j = 2; j <= N; ++j ){ if( vis[j] ) continue; if( d[j] == d[pos] ) mark[j] = true; } res += d[pos]; for( j = 2; j <= N; ++j ){ if( vis[j] || map[j][pos] == -1 ) continue; if( d[j] == -1 || d[j] >= map[j][pos] ){ if( mark[j] ) return false; if( d[j] == map[j][pos] ) v[j] = true; else v[j] = false; d[j] = map[j][pos]; } } } return true; } int main() { int T, i, j, x, y, c; scanf( "%d", &T ); while( T-- ){ scanf( "%d%d", &N, &M ); for( i = 1; i <= N; ++i ) for( j = 1; j <= N; ++j ) map[i][j] = -1; for( i = 0; i < M; ++i ){ scanf( "%d%d%d", &x, &y, &c ); if( map[x][y] == -1 || map[x][y] > c ) map[x][y] = map[y][x] = c; } if( prim()) printf( "%d/n", res ); else printf( "Not Unique!/n" ); } return 0; }