Texas's simple people are suffering a huge heat wave this summer!!! Their Texas horned cows are good to eat, but they are not very good at producing cream-rich dairy products. Farmer John at this time first, enjoy spirit, Shenxianshizu to Texas to carry a large number of nutritious cold milk, to alleviate the Texan people endure the pain of heat.
FJ has studied the route of transporting milk from Wisconsin to Texas. These routes include a starting point and an end point in a total of T (1 <= t <= 2,500) towns, conveniently labeled as 1 to T. In addition to the beginning and end of the field each town is connected by two bidirectional roads to at least two other towns. Each road has a pass fee (including fuel, toll, etc.).
Given a map that contains the C (1 <= C <= 6,200) strip directly connected to the road of 2 towns. Each road consists of the beginning of the road Rs, the end point Re (1 <= Rs <= t; 1 <= Re <= t), and the cost (1 <= Ci <=) . The minimum total cost for the town Te (1 <= te <= t) from the beginning of the town TS (1 <= TS <= t) to the end point .
This problem is used Dijkstra, AC also understand the meaning of the Dijkstra algorithm.
Variable function as its name, easy for beginners to learn together
1#include <stdio.h>2#include <stdlib.h>3#include <string.h>4 #defineMAX 10000005 intmap[2501][2501];//the previous dimension represents the starting point, and the latter one represents the terminating point6 intpointspassed[2501],shortestdist[2501];7 intT,c,ts,te;//t= Town number c= Road number ts= start town te= End Town8 intRs,re,ci;9 Ten intMain () One { A intI,j,nowmindist,topoint; -scanf"%d%d%d%d",&t,&c,&ts,&Te); - the for(i=1; i<=t;i++) - for(j=1; j<=t;j++) -map[i][j]=MAX; - + for(i=1; i<=c;i++) - { +scanf"%d%d%d",&rs,&re,&Ci); A if(Ci<=map[rs][re])//depending on the input there may be a cheaper way at { -map[rs][re]=Ci; -map[re][rs]=Ci; - } - } -map[ts][ts]=0; in - for(i=1; i<=t;i++)//Initialize to { +shortestdist[i]=Map[ts][i]; -pointspassed[i]=0; the } *pointspassed[ts]=1; $ Panax Notoginseng for(i=2; i<=t;i++) - { thenowmindist=MAX; +Topoint=ts;//each time from the starting point to find the shortest distance from the A //now it's just the shortest path known to this point . the + for(j=1; j<=t;j++) - if((!pointspassed[j]) && shortestdist[j]<nowmindist) $ { $Topoint=J; -nowmindist=Shortestdist[j]; - } the - //ingest this pointWuyipointspassed[topoint]=1; the for(j=1; j<=t;j++) - if((!pointspassed[j]) && Map[topoint][j]!=max)//extend the outward point from this point Wu if(Shortestdist[topoint] + map[topoint][j] <Shortestdist[j]) -Shortestdist[j]=shortestdist[topoint] +Map[topoint][j]; About } $ -printf"%d", Shortestdist[te]); - return 0; -}
Codevs 1557 heat wave