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 ci.

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,…,ln. The third line contains n integers r1,r2,…,rn. The fourth line contains n integers c1,c2,…,cn. (0≤li≤ri≤n,1≤ci≤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

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 ;
}

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