hdu5361 In Touch 神奇的dij，神奇的写法

In Touch

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 784    Accepted Submission(s): 217

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

 

dij算法求最短路的性质，与spfa不同已更新过的点的最短距离已经确定了。这道题用dij正合适，sq里是已更新过的点，se里是没更新过的点，为了使点不太多，每次更新se里的点时，把此点从se中删除，空间复杂度仅O(n).

#include<iostream>#include<cstdio>#include<set>#define inf 1000000000000000#define ll long longusing namespace std;const int maxn=200005;int n;int l[maxn];int r[maxn];ll c[maxn];ll dist[maxn];set<int>se;struct Node{    int id;    ll w;    Node(int a,ll b):id(a),w(b){}    bool operator<(const Node& u)const    {        if(w+c[id]==u.w+c[u.id]) {            return id<u.id;        }        return w+c[id]<u.w+c[u.id];    }};set<Node>sq;int main(){    int t;    scanf("%d",&t);    while(t--) {        se.clear();        sq.clear();        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("%I64d",&c[i]);        }        for(int i=1;i<=n;i++)            dist[i]=inf;        dist[1]=0;        sq.insert(Node(1,0));        for(int i=2;i<=n;i++) {            se.insert(i);        }        set<int>::iterator it;        set<int>::iterator it2;        while(!sq.empty()) {            Node u=*sq.begin();            sq.erase(sq.begin());            it=se.lower_bound(u.id-r[u.id]);            while(it!=se.end() && *it<=u.id-l[u.id]) {                if(dist[*it]>u.w+c[u.id]) {                    dist[*it]=u.w+c[u.id];                    sq.insert(Node(*it,dist[*it]));                    it2=it++;         //删除的巧妙写法                    se.erase(it2);                }            }            it=se.lower_bound(u.id+l[u.id]);            while(it!=se.end() && *it<=u.id+r[u.id]) {                if(dist[*it]>u.w+c[u.id]) {                    dist[*it]=u.w+c[u.id];                    sq.insert(Node(*it,dist[*it]));                    it2=it++;                    se.erase(it2);                }            }        }        printf("0");        for(int i=2;i<=n;i++) {            if(dist[i]==inf) {                printf(" -1");            }            else {                printf(" %I64d",dist[i]);            }        }        puts("");    }    return 0;}