The shortest path problem of Hangzhou electric 3790--(double-weighted minimum spanning tree)

Shortest path problem

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 17425 Accepted Submission (s): 5199

Problem description give you n points, M no edge, each side has a length D and spend P, give you the beginning of the end of T, the minimum distance from the output starting point to the end point and the cost, if the shortest distance there are multiple routes, the output cost is the least.

Input n,m, the number of the point is 1~n, then the M line, 4 numbers per line, A,b,d,p, indicates an edge between A and B, and its length is D, which costs p. The last line is two number s,t; start s, end point. N and m are 0 o'clock input end.
(1<n<=1000, 0<m<100000, s! = t)

The output line has two numbers, a minimum distance, and a cost.

Sample INPUT3 2 1 2 5 6 2 3 4 5 1 3 0 0

Sample Output9 11

SOURCE Zhejiang University Computer Postgraduate exam on the machine-2010

recommendnotonlysuccess | We have carefully selected several similar problems for you:2066 1217 2112 1142 1548 RE: The topic does not say that the cost is the least, so the distance and cost should be one by one corresponding;

1#include <cstdio>2#include <cstring>3 using namespacestd;4 Const intINF =0x3f3f3f;5 intmap[1010][1010], mpa[1010][1010], dis1[1010], dis2[1010], vis[1010];6 intI, J, M, N;7 voidDijkstra (intCRA)8 {9memset (Vis,0,sizeof(Vis));Ten      for(i=1; i<=m; i++) One     { ADis1[i] =Map[cra][i]; -Dis2[i] =Mpa[cra][i]; -     } theVIS[CRA] =1; -      for(i=1; i<m; i++) -     { -         intmin = INF, temp =CRA; +          for(j=1; j<=m; J + +) -         { +             if(!vis[j] && Dis1[j] <min) A             { attemp =J; -Min =Dis1[j]; -             } -         } -Vis[temp] =1; -          for(j=1; j<=m; J + +) in         { -             if(!vis[j] && dis1[j] > Dis1[temp] +Map[temp][j]) to             { +DIS1[J] = Dis1[temp] +Map[temp][j]; -DIS2[J] = Dis2[temp] +Mpa[temp][j]; the             } *             Else if(!vis[j] && dis1[j] = = Dis1[temp] + map[temp][j] && dis2[j] > dis2[temp]+Mpa[temp][j]) $             {Panax NotoginsengDIS2[J] = Dis2[temp] +Mpa[temp][j]; -             } the         } +     } A } the intMain () + { -      while(~SCANF ("%d%d", &m, &n), m|N) $     { $          for(i=1; i<=m; i++) -          for(j=1; j<=m; J + +) -         { theMap[i][j]= (i==j?0: INF); -Mpa[i][j]= (i==j?0: INF);Wuyi         } the         intA, B, D, p; -          for(i =1; I <= N; i++) Wu         { -scanf"%d %d%d%d", &a, &b, &d, &p); About         /*if (Map[a][b] > D)//pit.!  $ Map[a][b]=map[b][a]=d; - if (Mpa[a][b] > P) - mpa[a][b]=mpa[b][a]=p; */ -             if(Map[a][b] >d) A             { +map[a][b]=map[b][a]=D; thempa[a][b]=mpa[b][a]=p; -             } $         }     the         ints, t; thescanf"%d%d", &s, &t); the Dijkstra (s); theprintf"%d%d\n", Dis1[t], dis2[t]); -     } in     return 0; the}

The shortest path problem of Hangzhou electric 3790--(double-weighted minimum spanning tree)