Summer Training 1 e question paid roads (poj 3411) solution report

Question link: http://poj.org/problem? Id = 3411

N cities are connected by M. If a city is connected to cityia and cityb, if you have already passed cityc (may be equal to a), you can pay P. Otherwise, you must pay the r toll if you have not passed cityc before.

Note that a point can be repeated multiple times, that is, it may constitute a ring. Although the road is long, the road cost is low. This problem should be taken into account. There is also pruning: if the current toll is larger than the previous answer, it will be traced back.

Mincost is clearly a global variable, and it was declared again in main with multiple hands. After more than an hour, I found that = ....

1 # include <iostream> 2 # include <cstdio> 3 # include <cstdlib> 4 # include <cstring> 5 using namespace STD; 6 7 const int maxn = 1000 + 5; 8 int head [maxn]; 9 int point_vis [maxn]; 10 int n, m, mincost; 11 12 struct adjlist // neighbor list 13 {14 int B, C, p, r; 15 int next; 16} node [maxn]; 17 18 void DFS (INT next, int cost) 19 {20 if (point_vis [next]> N | cost> = mincost) // If a point is passed more than N, or if the current cost is greater than the minsum obtained previously, it will be traced back. 21 return; 22 if (next = N) 23 {24 mincost = cost; 25 return; 26} 27 for (INT I = head [next]; I! =-1; I = node [I]. next) 28 {29 int v = node [I]. b; 30 point_vis [node [I]. b] ++; 31 if (point_vis [node [I]. c]> = 1) // if you have accessed this intermediate point C, you can pay the fee of p32 DFS (node [I]. b, cost + node [I]. p); 33 else // previously this point has been taken over, and you can only pay the r fee 34 DFS (node [I]. b, cost + node [I]. r); 35 point_vis [node [I]. b] --; 36} 37} 38 39 int main () 40 {41 while (scanf ("% d", & N, & M )! = EOF) 42 {43 int T1, CNT = 0; 44 memset (Head,-1, sizeof (head); 45 for (INT I = 0; I <m; I ++) 46 {47 scanf ("% d", & T1); 48 node [CNT]. next = head [T1]; 49 scanf ("% d", & node [CNT]. b, & node [CNT]. c, & node [CNT]. p, & node [CNT]. r); 50 head [T1] = CNT ++; 51} 52 memset (point_vis, 0, sizeof (point_vis); 53 mincost = maxn; 54 point_vis [1] ++; 55 DFS (1, 0); 56 If (mincost = maxn) 57 printf ("impossible \ n "); 58 else59 printf ("% d \ n", mincost); 60} 61 Return 0; 62}