sample ssi %d 兩個 int chm das enter needed

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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 from left to right. The distance between two adjacent soda is 1 meter. Every soda has a teleporter. The teleporter of -th soda can teleport to the soda whose distance between -th soda is no less than and no larger than . The cost to use -th soda‘s teleporter is .

The -st soda is their leader and he wants to know the minimum cost needed to reach -th soda .

Input There are multiple test cases. The first line of input contains an integer , indicating the number of test cases. For each test case:

The first line contains an integer , the number of soda.
The second line contains integers . The third line contains integers . The fourth line contains integers .
Output For each case, output integers where -th integer denotes the minimum cost needed to reach -th soda. If -st soda cannot reach -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
HintIf 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：把一個集合的點看做是一個點，這樣就能夠用djstra算法做了。然後因為每一個點最多標記一次最短路，用set維護一個點集合。

當最短路找到一個一個集合的時候，把這個集合裏還存在的點都取出就可以。取出後。每一個點又能夠去兩個集合。

再向保存最短路的set裏更新集合信息就可以。具體看代碼。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
using namespace std;
#define maxn 200007
#define ll long long

int 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;
}

hdu 5361 2015多校聯合訓練賽#6 最短路