题目链接：
题目描述：
有n个传送机排成一排，编号从1到n，每个传送机都可以把自己位置的东西传送到距离自己[l, r]距离的位置，并且花费c，问从1号传送机到其他传送机的最小花费是多少？？
解题思路：
看了题解发现是dijkstra(求最短单源路径)+并查集(优化)，刚开始的时候并不知道怎么用并查集优化，然后就提交了一个没有并查集的代码，果断TLE。然后开始搜代码，突然惊叹真是优化的太美妙。因为每次从队列中取出的都是最优的，所以更新dis的时候可以直接把这些点合并，下次再遇到这些点的时候没必要更新就直接跳过。(在这里要特别谢谢为我debug快要疯了的天I火聚聚)
1 #include <cmath> 2 #include <queue> 3 #include <cstdio> 4 #include <cstring> 5 #include <iostream> 6 #include <algorithm> 7 using namespace std; 8 typedef long long LL; 9 const LL INF = 1LL<<62; 10 const int maxn = 200005; 11 struct node 12 { 13 int x; 14 LL dis; 15 node (int a, LL b):x(a),dis(b){} 16 friend bool operator < (node a, node b) 17 { 18 return a.dis > b.dis; 19 } 20 }; 21 int l[maxn], r[maxn], c[maxn], father[maxn], n; 22 LL dis[maxn]; 23 int Find (int x) 24 { 25 if (x != father[x]) 26 father[x] = Find (father[x]); 27 return father[x]; 28 } 29 void Merge (int x, int y) 30 { 31 int px = Find(x); 32 int py = Find(y); 33 if (px == py) 34 return ; 35 father[px] = py; 36 } 37 void dijkstra (int x) 38 { 39 priority_queue <node> Q; 40 dis[x] = c[x]; 41 Q.push(node(x, dis[x])); 42 while (!Q.empty()) 43 { 44 int p = Q.top().x; 45 Q.pop(); 46 for (int i=-1; i<=1; i+=2) 47 { 48 int L = p + l[p]*i; 49 int R = p + r[p]*i; 50 if (L > R) 51 swap(L, R); 52 L = max (L, 1); 53 R = min (R, n); 54 for (int j=L; j<=R; j++) 55 { 56 j = Find (j); 57 if ( j > R) 58 break; 59 if (dis[j] > dis[p] + c[j]) 60 { 61 dis[j] = dis[p] + c[j]; 62 Q.push(node(j, dis[j])); 63 } 64 Merge (j, j+1); 65 } 66 } 67 } 68 } 69 int main () 70 { 71 int t; 72 scanf ("%d", &t); 73 while (t --) 74 { 75 scanf ("%d", &n); 76 for (int i=1; i<=n; i++) 77 scanf ("%d", &l[i]); 78 for (int i=1; i<=n; i++) 79 scanf ("%d", &r[i]); 80 for (int i=1; i<=n; i++) 81 scanf ("%d", &c[i]); 82 for (int i=0; i<=n+1; i++) 83 { 84 father[i] = i; 85 dis[i] = INF; 86 } 87 dijkstra (1); 88 for (int i=1; i<=n; i++) 89 { 90 if (dis[i] == INF) 91 dis[i] = c[i] - 1; 92 printf ("%lld%c", dis[i]-c[i], i==n?' ':' '); 93 } 94 } 95 return 0; 96 }