1031. Campus

TAG 最短路 Dijkstra

经典题目，直接用Dijkstra就行，只是注意要把字符串映射成编号，我用STL的map容器做的。还有节点数最多是200

/* source code of submission 424628, Zhongshan University Online Judge System */ #include <stdio.h> #include <map> #include <string> #include <memory.h> using namespace std; char s1[110],s2[110]; int g[200][200]; int t,n; int idx1,idx2,len; int counter; map<string,int> node; int findIndex( char *s ) { if ( node.count(s)==0 ) { ++counter; node[s]=counter; return counter; } else { return node[s]; } } int Dijkstra(int s,int t) { int dist[200]; bool visited[200]; if (s==t) return 0; memset(visited, false, sizeof(visited) ); memset(dist, -1, sizeof(dist) ); dist[s]=0; visited[s]=true; for (int i=0; i<=counter; ++i) { if ( i!=s ) { dist[i]=g[s][i]; } } for (int k=1; k<=counter; ++k) { int mind=90000; int idx=-1; for (int i=0; i<=counter; ++i) { if ( !visited[i] && dist[i]!=-1 && dist[i]<mind ) { mind=dist[i]; idx=i; } } visited[idx]=true; if ( idx==t ) { return dist[t]; } for (int i=0; i<=counter; ++i) { if ( !visited[i] && g[idx][i]!=-1 ) { if (dist[i]==-1 || dist[i]> dist[idx]+g[idx][i] ) { dist[i]=dist[idx]+g[idx][i]; } } } } return -1; } int main(int argc, char *argv[]) { scanf("%d", &t); while ( t-- ) { scanf("%d", &n); node.clear(); counter=-1; memset(g, -1, sizeof(g) ); for (int i=0; i<n; ++i) { scanf("%s%s%d", s1, s2, &len ); idx1=findIndex(s1); idx2=findIndex(s2); g[idx1][idx2]=len; g[idx2][idx1]=len; } scanf("%s%s",s1, s2); printf("%d/n", Dijkstra( findIndex(s1), findIndex(s2) ) ); } return 0; }