Travel Planning (self-written Floyd code + DIJ algorithm not learned)

Source: Internet
Author: User

With a road map of self-driving, you will know the length of the highway between cities and the tolls of the road to be charged. Now you need to write a program to help visitors find a shortest route between the origin and the destination. If you have several paths that are the shortest, you need to output the cheapest path.

Input format:

Input Description: The 1th line of the input data gives 4 positive integersNMSD, whereN2) is the number of cities, by the way assuming the city's number is 0~ (n? 1); m is the number of highways; S is the city number of the departure point; d is the city number of the destination. Subsequent m line, each line gives the information of a highway, respectively: City 1, City 2, highway length, charge amount, the middle with a space separate, The numbers are integers and do not exceed 500. The input guarantees the existence of the solution.

Output format:

In a row, the length of the output path and the total charge, the numbers are separated by a space, the output end can not have extra space.

Input Sample:
4 5 0 3 0 1 1  - 1 3 2  - 0 3 4 Ten 0 2 2  - 2 3 1  -
Sample output:
3  +

Floyd Code

#include <iostream>#include<string>#include<cmath>using namespacestd;#defineMAX 1e9struct {    intl, S;} a[ -][ -];intMain () {intN, M, S, D,i,j,k,x,y,ll,ss; CIN>> N >> M >> S >>D;  for(i =0; I < n;i++)     for(j =0; J < N; J + +)    {        if(i = =j) {A[I][J].L=0; A[i][j].s=0; }        Else{A[I][J].L=MAX; A[i][j].s=MAX; }    }     while(m--) {cin>> x >> y >> ll >>SS; A[X][Y].L=ll; A[Y][X].L=ll; A[y][x].s=SS; A[x][y].s=SS; }     for(k =0; K < n;k++)     for(i =0; I < n;i++)     for(j =0; J < n;j++)    if(A[I][J].L&GT;A[I][K].L + a[k][j].l | | a[i][j].l = = A[I][K].L + A[k][j].l&&a[i][j].s>a[i][k].s +a[k][j].s) {A[I][J].L= a[i][k].l+A[K][J].L; A[i][j].s= A[i][k].s +A[k][j].s; } cout<< A[S][D].L <<" "<<A[s][d].s;}
Floyd algorithm (alas, the ladder will not learn before the game)
#include <stdio.h>#defineMAX 100000typedefstruct{    intweight; intCost ;} Graph;intn,m,s,d;graph g[ -][ -];intdis[ -];intcost[ -];intflag[ -];voidDijkstra (intv) {    intMin,i,j,pos;  for(i=0; i<n;i++)    {        if(g[v][i].weight>0) {Dis[i]=G[v][i].weight; Cost[i]=G[v][i].cost; }} Flag[v]=1;  for(i=0; i<n;i++) {min=MAX; POS=v;  for(j=i;j<n;j++)        {            if(flag[j]!=1&&dis[j]<min &&dis[j]>0) {min=Dis[j]; POS=J; }} Flag[pos]=1;  for(j=0; j<n;j++)        {            if(flag[j]!=1&AMP;&AMP;DIS[POS]+G[POS][J].WEIGHT&LT;DIS[J] &&g[pos][j].weight>0&&dis[j]>0) {Dis[j]=dis[pos]+G[pos][j].weight; COST[J]=cost[pos]+G[pos][j].cost; }            Else if(Dis[pos]+g[pos][j].weight==dis[j] &&cost[j]>cost[pos]+g[pos][j].cost) {Cost[j]=cost[pos]+G[pos][j].cost; }}} printf ("%d%d\n", Dis[d],cost[d]); }main () {intA,b,l,c; inti,j; scanf ("%d%d%d%d",&n,&m,&s,&D);  for(i=0; i<m;i++) {scanf ("%d%d%d%d",&a,&b,&l,&c); G[a][b].weight=g[b][a].weight=l; G[a][b].cost=g[b][a].cost=B; } Dijkstra (S);}

Travel Planning (self-written Floyd code + DIJ algorithm not learned)

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.