Hdu 5361 In Touch (dijkatrs+优先队列)

题目链接：

　　Hdu 5361  In Touch

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)

 #include <cmath>
 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 const LL INF = 1LL<<;
 const int maxn = ;
 struct node
 {
     int x;
     LL dis;
     node (int a, LL b):x(a),dis(b){}
     friend bool operator < (node a, node b)
     {
         return a.dis > b.dis;
     }
 };
 int l[maxn], r[maxn], c[maxn], father[maxn], n;
 LL dis[maxn];
 int Find (int x)
 {
     if (x != father[x])
         father[x] = Find (father[x]);
         return father[x];
 }
 void Merge (int x, int y)
 {
     int px = Find(x);
     int py = Find(y);
     if (px == py)
         return ;
     father[px] = py;
 }
 void dijkstra (int x)
 {
     priority_queue <node> Q;
     dis[x] = c[x];
     Q.push(node(x, dis[x]));
     while (!Q.empty())
     {
         int p = Q.top().x;
         Q.pop();
         for (int i=-; i<=; i+=)
         {
             int L = p + l[p]*i;
             int R = p + r[p]*i;
             if (L > R)
                 swap(L, R);
             L = max (L, );
             R = min (R, n);
             for (int j=L; j<=R; j++)
             {
                 j = Find (j);
                 if ( j > R)
                     break;
                 if (dis[j] > dis[p] + c[j])
                 {
                     dis[j] = dis[p] + c[j];
                     Q.push(node(j, dis[j]));
                 }
                 Merge (j, j+);
             }
         }
     }
 }
 int main ()
 {
     int t;
     scanf ("%d", &t);
     while (t --)
     {
         scanf ("%d", &n);
         for (int i=; i<=n; i++)
             scanf ("%d", &l[i]);
         for (int i=; i<=n; i++)
             scanf ("%d", &r[i]);
         for (int i=; i<=n; i++)
             scanf ("%d", &c[i]);
         for (int i=; i<=n+; i++)
             {
                 father[i] = i;
                 dis[i] = INF;
             }
             dijkstra ();
         for (int i=; i<=n; i++)
         {
             if (dis[i] == INF)
                 dis[i] = c[i] - ;
             printf ("%lld%c", dis[i]-c[i], i==n?'\n':' ');
         }
     }
     return ;
 }