hdu 5361 2015多校联合训练赛#6 最短路

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Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 67    Accepted Submission(s): 11

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++.
 

 

Source

2015 Multi-University Training Contest 6

求最短路：把一个集合的点看做是一个点，这样就可以用djstra算法做了。然后由于每个点最多标记一次最短路，用set维护一个点集合。

当最短路找到一个一个集合的时候，把这个集合里还存在的点都取出即可。取出后，每个点又可以去两个集合，

再向保存最短路的set里更新集合信息即可。详细看代码。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<set>using namespace std;#define maxn 200007#define ll long longint lp[maxn],rp[maxn];ll cosw[maxn];ll dist[maxn];set<int> haha;struct Node{    int id;    ll cost;};bool operator < (Node a,Node b){    if(a.cost == b.cost) return a.id < b.id;    return a.cost < b.cost;}set<Node> mind;int main(){    int t,n;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(int i = 0;i < n; i++)            scanf("%d",&lp[i]);        for(int i = 0;i < n; i++)            scanf("%d",&rp[i]);        for(int i = 0;i < n; i++)            scanf("%d",&cosw[i]);        haha.clear();        mind.clear();        memset(dist,-1,sizeof(dist));        dist[0] = 0;        Node x,y;        x.id = 0;        x.cost = cosw[0];        mind.insert(x);        for(int i = 1;i < n; i++)            haha.insert(i);        set<int>::iterator it,it2;        while(mind.size() > 0){            x = *mind.begin();            mind.erase(mind.begin());            it = haha.lower_bound(x.id - rp[x.id]);            while(it != haha.end()  && *it <= x.id - lp[x.id]){                y.id = *it;                y.cost = x.cost + cosw[y.id];                dist[y.id] = x.cost;                mind.insert(y);                it2 = it++;                haha.erase(it2);            }            it = haha.lower_bound(x.id + lp[x.id]);            while(it != haha.end()  && *it <= x.id + rp[x.id]){                y.id = *it;                y.cost = x.cost + cosw[y.id];                dist[y.id] = x.cost;                mind.insert(y);                it2 = it++;                haha.erase(it2);            }        }        for(int  i = 0;i < n; i++){            if(i) printf(" ");            printf("%I64d",dist[i]);        }        printf("\n");    }    return 0;}