HDu 5361

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Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1609    Accepted Submission(s): 435

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

152 0 0 0 13 1 1 0 51 1 1 1 1

 

Sample Output

0 2 1 1 -1
Hint
If you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.
 

 

#include <iostream>#include <stdio.h>#include <string.h>#include <queue>#include <cmath>#include <algorithm>using namespace std;const int maxn=2*10e5+10;const __int64 inf=(1LL<<62);__int64 dist[maxn];int l[maxn],r[maxn],c[maxn],pre[maxn];struct Node{    int x;    __int64 dist;    friend bool operator < (const Node &a,const Node &b)    {        return a.dist>b.dist;    }};priority_queue <Node> q;int fin(int x){    int r=x;    while(r!=pre[r])    {        r=pre[r];    }    int k=x;    while(pre[k]!=r)    {        int j=pre[k];        pre[k]=r;        k=j;    }    return r;}void join(int x,int y){    int fx=fin(x);    int fy=fin(y);    if(fx!=fy)        pre[fx]=fy;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&l[i]);        for(int i=1;i<=n;i++)            scanf("%d",&r[i]);        for(int i=1;i<=n;i++)            scanf("%d",&c[i]);        for(int i=1;i<=n+1;i++)        {            dist[i]=inf;            pre[i]=i;        }        Node a;        dist[1]=c[1];        a.x=1,a.dist=0;        q.push(a);        while(!q.empty())        {            Node b=q.top();            q.pop();            int num=b.x;            for(int i=-1;i<=1;i+=2)            {                int L=num+l[num]*i;                int R=num+r[num]*i;                if(L>R) swap(L,R);                if(L>n||R<=0)continue;                L=max(L,1),R=min(R,n);                for(int k=L;k<=R;k++)                {                    k=fin(k);                    if(k>R)break;                    if(dist[k]>dist[num]+c[k])                    {                        dist[k]=dist[num]+c[k];                        Node c;                        c.x=k,c.dist=dist[k];                        q.push(c);                    }                    join(k,k+1);                }            }        }        printf("0");        for(int i=2;i<=n;i++)        {            if(dist[i]==inf)                printf(" -1");            else                printf(" %I64d",dist[i]-c[i]);        }        printf("\n");    }    return 0;}