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 = 200010;
struct node{
    int id;
    LL cost;
    node(int x = 0, LL y = 0){
        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,-1,sizeof d);
        p.clear();
        q.clear();
        d[0] = 0;
        for(int i = 0; i < n; ++i){
            scanf("%d",L + i);
            if(i) p.insert(i);
        }
        for(int i = 0; i < n; ++i)
            scanf("%d",R + i);
        for(int i = 0; i < n; ++i)
            scanf("%I64d",w + i);
        q.insert(node(0,w[0]));
        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 = 0; i < n; ++i)
            printf("%I64d%c",d[i],i + 1 == n?'
':' ');
    }
    return 0;
}