2015 Multi-University Training Contest 6 hdu 5361 In Touch

In Touch

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 578    Accepted Submission(s): 160

Problem Description

There are n soda living in a straight line. soda are numbered by 1,2,…,n from left to right. The distance between two adjacent soda is 1 meter. Every soda has a teleporter. The teleporter of i-th soda can teleport to the soda whose distance between i-th soda is no less than li and no larger than ri. The cost to use i-th soda's teleporter is c_i.

The 1-st soda is their leader and he wants to know the minimum cost needed to reach i-th soda (1≤i≤n).

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤2×10⁵), the number of soda.
The second line contains n integers l₁,l₂,…,l_n. The third line contains n integers r₁,r₂,…,r_n. The fourth line contains n integers c₁,c₂,…,c_n. (0≤l_i≤r_i≤n,1≤c_i≤10⁹)

Output
For each case, output n integers where i-th integer denotes the minimum cost needed to reach i-th soda. If 1-st soda cannot reach i-the soda, you should just output -1.

Sample Input

1

5

2 0 0 0 1

3 1 1 0 5

1 1 1 1 1

 

Sample Output

0 2 1 1 -1

 

Source

2015 Multi-University Training Contest 6 1009-I

解题：利用set可以二分，进行dijkstra，思路是学习这位大神的，确实很赞，很厉害。。。。很奇妙

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 struct node{
     int id;
     LL cost;
     node(int x = , LL y = ){
         id = x;
         cost = y;
     }
     bool operator<(const node &t)const{
         if(cost == t.cost) return id < t.id;
         return cost < t.cost;
     }
 };
 set<int>p;
 set<node>q;
 int L[maxn],R[maxn],n;
 LL w[maxn],d[maxn];
 int main(){
     int kase;
     scanf("%d",&kase);
     while(kase--){
         scanf("%d",&n);
         memset(d,-,sizeof d);
         p.clear();
         q.clear();
         d[] = ;
         for(int i = ; i < n; ++i){
             scanf("%d",L + i);
             if(i) p.insert(i);
         }
         for(int i = ; i < n; ++i)
             scanf("%d",R + i);
         for(int i = ; i < n; ++i)
             scanf("%I64d",w + i);
         q.insert(node(,w[]));
         while(q.size()){
             node cur = *q.begin();
             q.erase(q.begin());
             auto it = p.lower_bound(cur.id - R[cur.id]);
             while(it != p.end() && *it <= cur.id - L[cur.id]){
                 d[*it] = cur.cost;
                 q.insert(node(*it,cur.cost + w[*it]));
                 p.erase(it++);
             }
             it = p.lower_bound(cur.id + L[cur.id]);
             while(it != p.end() && *it <= cur.id + R[cur.id]){
                 d[*it] = cur.cost;
                 q.insert(node(*it,cur.cost + w[*it]));
                 p.erase(it++);
             }
         }
         for(int i = ; i < n; ++i)
             printf("%I64d%c",d[i],i +  == n?'\n':' ');
     }
     return ;
 }