Topic : https://www.luogu.org/problemnew/show/P1339
Meaning of the questions : Given a map, ask the shortest beginning to the end.
Ideas : dijkstra board title.
Long time no write most short-circuited. Step dijkstra's summarize it.
d represents the current array for the shortest path, vis array to mark whether the current point is already set in the shortest.
Each time find a minimum of node d, he said he has been unable to shorter and took him to the collection, use him to update the other nodes. We made a total of n-1 times.
1 #include<cstdio> 2 #include<cstdlib> 3 #include<map> 4 #include<set> 5 #include<cstring> 6 #include<algorithm> 7 #include<vector> 8 #include<cmath> 9 #include<stack> 10 #include<queue> 11 #include<iostream> 12 13 #define inf 0x3f3f3f3f 14 using namespace std; 15 typedef long long LL; 16 typedef pair<int, int> pr; 17 18 int t, c, ts, te; 19 const int maxn = 2500; 20 const int maxm = 6205; 21 struct edge{ 22 int to, nxt, cost; 23 }e[maxm * 2]; 24 int head[maxn], tot; 25 26 void add(int x, int y, int w) 27 { 28 e[++tot].to = y; 29 e[tot].cost = w; 30 e[tot].nxt = head[x]; 31 head[x] = tot; 32 e[++tot].to = x; 33 e[tot].cost = w; 34 e[tot].nxt = head[y]; 35 head[y] = tot; 36 } 37 38 bool vis[maxn]; 39 int d[maxn]; 40 void dijkstra() 41 { 42 memset(d, 0x3f, sizeof(d)); 43 for(int i = head[ts]; i; i = e[i].nxt){ 44 d[e[i].to] = e[i].cost; 45 } 46 vis[ts] = true; 47 d[ts] = 0; 48 for(int i = 1; i < t; i++){ 49 int min = inf, min_id; 50 for(int j = 1; j <= t; j++){ 51 if(d[j] < min && !vis[j]){ 52 min = d[j]; 53 min_id = j; 54 } 55 } 56 vis[min_id] = true; 57 for(int i = head[min_id]; i; i = e[i].nxt){ 58 if(d[e[i].to] > min + e[i].cost){ 59 d[e[i].to] = min + e[i].cost; 60 } 61 } 62 } 63 } 64 65 int main() 66 { 67 scanf("%d%d%d%d", &t, &c, &ts, &te); 68 for(int i = 0; i < c; i++){ 69 int rs, re, ci; 70 scanf("%d%d%d", &rs, &re, &ci); 71 add(rs, re, ci); 72 } 73 74 dijkstra(); 75 printf("%d\n", d[te]); 76 }