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Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 178    Accepted Submission(s): 44

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

大致思路： 从第一个点开始更新， 更新过的点缩成一个点， 因为在Dijkstra里 每次取出的都是最小的distance， 所以更新过的点 后面肯定不需要再次更新。更新一个点后加入堆中， 因为通过这个点可能更新别的点。

#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <queue>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <functional>
#include <algorithm>
using namespace std;
const int MAXN = 2e5+10;
typedef long long LL;
typedef pair <LL, int>pli;
const LL inf  = 1LL << 60;
LL L[MAXN], R[MAXN], cost[MAXN], dis[MAXN];
int dsu[MAXN];
void init (){
    for (int i = 0 ; i< MAXN; i++){
        dis[i] = inf;
        dsu[i] = i;
    }
}
int find (int s){
    return dsu[s] = (dsu[s] == s ? s : find(dsu[s]));
}
int main() {
    int n, T;
    scanf ("%d", &T);
    while (T--) {
        init();
        scanf ("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf ("%I64d", L+i);
        }
        for (int i = 1; i <= n; i++) {
            scanf ("%I64d", R+i);
        }
        for (int i = 1; i <= n; i++) {
            scanf ("%I64d", cost+i);
        }
        init();
        dis[1] = cost[1];
        priority_queue<pli, vector<pli>, greater<pli> >Q;
        Q.push(make_pair(0LL, 1));
        while (!Q.empty()){
            pli tmp = Q.top();
            Q.pop();
            int u = tmp.second;
            for (int i = -1; i <= 1; i += 2){
                int lf = L[u] * i + u;
                int rg = R[u] * i + u;
                if (lf > rg){
                    swap(lf, rg);
                }
                lf = max(lf, 1);
                lf = min(lf, n + 1);
                if (lf > rg){
                    continue;
                }
                for (int v = lf; ; v ++){
                    v = find(v);
                    if (v <= 0 || v > n || v > rg){
                        break;
                    }
                    if (dis[v] > dis[u] + cost[v]){
                        dis[v] = dis[u] + cost[v];
                        Q.push(make_pair(dis[v], v));
                    }
                    dsu[find(v)] = find(v + 1);
                }
            }
        }
        printf("0");
        for (int i = 2; i <= n; i++) {
            printf(" %I64d", dis[i] != inf ? dis[i] - cost[i] : -1);
        }
        printf("\n");
    }
    return 0;
}