A strange lift

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题意：有n层楼层，现在在每一层有两个按钮，分别为up和down，按动按钮时，可以向上或向下跳动num[ i ]层；问能否以最少的次数从A到B层，不能输出-1；

解析：构图，将从i层到按动按钮后跳转的楼层，看作连通状态，赋值为1，这样就转换成单源最短路问题；

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int maxn = 500;const int Max = 0x3f3f3f3f;int num[ maxn ], mapp[ maxn ][ maxn ], dis[ maxn ], vis[ maxn ];int n, start, end;void Dijkstra( int start ){memset( vis, 0, sizeof( vis ) );vis[ start ] = 1;for( int i = 1; i <= n; ++i ){dis[ i ] = mapp[ start ][ i ];}/*for( int i = 0; i <= n; ++i ){cout << dis[ i ] << endl;}*/dis[ start ] = 0;for( int i = 1; i <= n; ++i ){int temp = Max, k;for( int j = 1; j <= n; ++j ){if( !vis[ j ] && temp > dis[ j ] ){temp = dis[ k = j ];}}if( temp == Max )break;vis[ k ] = 1;for( int j = 1; j <= n; ++j ){if( !vis[ j ] && dis[ j ] > dis[ k ] + mapp[ k ][ j ] ){dis[ j ] = dis[ k ] + mapp[ k ][ j ];//vis[ j ] = 1;}}}/*for( int i = 0; i <= n; ++i ){cout << dis[ i ] << endl;}*/}int main(){while( scanf( "%d", &n ) != EOF ){if( !n )break;scanf( "%d%d", &start, &end );for( int i = 0; i <= n; ++i ){for( int j = 0;j <= n; ++j ){mapp[ i ][ j ] = Max;}}for( int i = 1; i <= n; ++i ){scanf( "%d", &num[ i ] );}for( int i = 1; i <= n; ++i ){if( i + num[ i ] <= n ){mapp[ i ][ i + num[ i ] ] = 1;}if( i - num[ i ] >= 1 ){mapp[ i ][ i - num[ i ] ] = 1;}}Dijkstra( start );if( dis[ end ] == Max )puts( "-1" );else{printf( "%d\n", dis[ end ] );}}return 0;}