iss pst dir 搜索 原因 ont \n rate vector

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Remmarguts‘ Date

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions:35025		Accepted: 9467

Description

"Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks‘ head, he told them a story.

"Prince Remmarguts lives in his kingdom UDF – United Delta of Freedom. One day their neighboring country sent them Princess Uyuw on a diplomatic mission."

"Erenow, the princess sent Remmarguts a letter, informing him that she would come to the hall and hold commercial talks with UDF if and only if the prince go and meet her via the K-th shortest path. (in fact, Uyuw does not want to come at all)"

Being interested in the trade development and such a lovely girl, Prince Remmarguts really became enamored. He needs you - the prime minister‘s help!

DETAILS: UDF‘s capital consists of N stations. The hall is numbered S, while the station numbered T denotes prince‘ current place. M muddy directed sideways connect some of the stations. Remmarguts‘ path to welcome the princess might include the same station twice or more than twice, even it is the station with number S or T. Different paths with same length will be considered disparate.

Input

The first line contains two integer numbers N and M (1 <= N <= 1000, 0 <= M <= 100000). Stations are numbered from 1 to N. Each of the following M lines contains three integer numbers A, B and T (1 <= A, B <= N, 1 <= T <= 100). It shows that there is a directed sideway from A-th station to B-th station with time T.

The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).

Output

A single line consisting of a single integer number: the length (time required) to welcome Princess Uyuw using the K-th shortest path. If K-th shortest path does not exist, you should output "-1" (without quotes) instead.

Sample Input

2 2
1 2 5
2 1 4
1 2 2

Sample Output

14

思路

或許將其分在最短路不是特別合適，但是暫時就這樣吧
此題為第k短路，用到了a*算法
這是一個神奇的算法。用它來求第k小的話，就是用它不斷搜索，即使遇到了終點也不停止。
首先我們知道，這個東西本來就是求最短路的一個算法，所以第一次遇到終點，一定是最短路。遇到終點後，如果不停止，那麽接下來，在優先隊列裏面第二靠近終點的肯定會及時補充上。總之，在這個算法裏面，
沒有vis數組的限制，又將耗費最小的放在前面，所以第二個到達終點的路徑就是第二短的了。
或許這樣太過於抽象，但是你只需要記住一點，那就是在這個題目裏面的A*算法，它的估計值是精確的，因為我們估計值的來源，是對終點反向圖的Dijkstra求出來的。正是這個原因，這個算法跑到的每一次終點
，都是準確的第幾小。

代碼
 

#include<iostream>
#include<vector>
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
const int inf = 2100000000;
vector<int>u[200024],w[200024];
vector<int>ux[200024],wx[200024];
bool book[200024];
int dis[200024];
int n,m,s,t,k;
struct node
{
    int num;
    int dis;
    bool operator<(const node x)const
    {
        return dis>x.dis;
    }
};

struct aa
{
    int num;
    int g,f;
    bool operator<(const aa x)const
    {
        if(x.f==f){return x.g<g;}
        return x.f<f;
    }
};

void Dijkstra()
{
    node exa;
    fill(dis,dis+n+5,inf);
    priority_queue<node>q;
    q.push(node{t,0});
    dis[t]=0;
    while(!q.empty()){
        exa=q.top();q.pop();
        if(book[exa.num]){continue;}
        book[exa.num]=true;
        int siz=ux[exa.num].size();
        for(int i=0;i<siz;i++){
            if(dis[ux[exa.num][i]]>dis[exa.num]+wx[exa.num][i]){
                dis[ux[exa.num][i]]=dis[exa.num]+wx[exa.num][i];
                q.push(node{ux[exa.num][i],dis[ux[exa.num][i]]});
            }
        }
    }
}

int A_star()
{
    aa exa;
    if(dis[s]==inf){return -1;}
    priority_queue<aa>q;
    int cnt = 0;
    if(s==t){k++;}

    int r,g,f;
    g=0;f=g+dis[s];
    q.push(aa{s,g,f});
    while(!q.empty()){
        exa= q.top();
        q.pop();
        r=exa.num;
        if(r==t){cnt++;}
        if(cnt==k){return exa.g;}
        int siz=u[r].size();
        for(int i=0;i<siz;i++){
            g=exa.g+w[r][i];
            f=g+dis[u[r][i]];
            q.push(aa{u[r][i],g,f});
        }
    }
    return -1;
}

int main()
{
    scanf("%d%d",&n,&m);
    int x,y,z;
    for(int i=1;i<=m;i++){
        scanf("%d%d%d",&x,&y,&z);
        u[x].push_back(y);
        w[x].push_back(z);
        ux[y].push_back(x);
        wx[y].push_back(z);
    }
    scanf("%d%d%d",&s,&t,&k);
    Dijkstra();
    printf("%d\n",A_star());
    return 0;
}

　　

POJ 2449 Remmarguts' Date