Sicily 4478. Walk This Way

第三届新手赛AC的第三题，也是最后一题（这届新手赛恰好过3题的队伍从第8名一直排到第37名，我队第33名，主要是经验缺乏）。开始时没有注意到是无向图，而且错误地使用了memset函数，导致WA。问题求的是从s点到t点最少要走多少路，解决方案可以用Floyd最短路径算法，也可以是Dijkstra单源最短路径算法。当时做这道题时，鉴于前者没有想到，后者不会实现，最后是通过对宽搜的改进AC的。大致思路是把visited这个bool数组改成int数组，让其携带访问对应点的最短距离信息，宽搜时再判断访问改点时是否会产生更短距离的路径来决定是否进行再访问。虽然当时的做法代码较为简便，但毕竟有点非主流；而Floyd算法求出了每两个点之间的最短路径，却有很多都是我们不关心（我们只关心s点到t点）的。综合而言，时间效率最高、最主流的算法应该就是Dijkstra单源最短路径算法，后来（新手赛结束后3天）在课堂上学习后写出该代码。

Run Time: 0sec

Run Memory: 304KB

Code length: 1395Bytes

Submit Time: 2011-12-21 23:51:11

// Problem#: 4478
// Submission#: 1107779
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/ // All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <cstdio>
using namespace std;

int main()
{
int T;
int i, j;
int N, M;
int a, b, c, d;
int s, t;
int min, add;
bool included[ 100 ];
int walk[ 100 ];
int dist[ 100 ][ 100 ];

scanf( "%d", &T );
while ( T-- ) {
scanf( "%d%d", &N, &M );
for ( i = 0; i < N; i++ ) {
included[ i ] = false;
for ( j = 0; j < N; j++ )
dist[ i ][ j ] = 10001;
}

for ( i = 0; i < M; i++ ) {
scanf( "%d%d%d%d", &a, &b, &c, &d );
dist[ a ][ b ] = ( d == 1 ? 0: c );
dist[ b ][ a ] = ( d == 1 ? 0: c );
}
scanf( "%d%d", &s, &t );
for ( i = 0; i < N; i++ )
walk[ i ] = dist[ s ][ i ];

add = s;
included[ s ] = true;
for ( j = 1; !included[ t ] && j <= N; j++ ) {
min = 10001;
for ( i = 0; i < N; i++ ) {
if ( !included[ i ] && walk[ i ] < min ) {
add = i;
min = walk[ i ];
}
}
included[ add ] = true;
for ( i = 0; i < N; i++ ) {
if ( !included[ i ] && min + dist[ add ][ i ] < walk[ i ] )
walk[ i ] = min + dist[ add ][ i ];
}
}
printf( included[ t ] ? "%d\n": "-1\n", walk[ t ] );
}

return 0;

}