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

1 5 2 0 0 0 1 3 1 1 0 5 1 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
#include
#include
#include
#include
using namespace std;
#define maxn 200007
#define ll long long

int lp[maxn],rp[maxn];
ll cosw[maxn];
ll dist[maxn];


set 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 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::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;
}