2763: [JLOI2011]飞行路线

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://www.lydsy.com/JudgeOnline/problem.php?id=2763

Description

Input

对于30%的数据,2<=n<=50,1<=m<=300,k=0;

对于50%的数据,2<=n<=600,1<=m<=6000,0<=k<=1;

对于100%的数据,2<=n<=10000,1<=m<=50000,0<=k<=10.

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <bitset>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 100010

#define eps 1e-9

int Num;

//const int inf=0x7fffffff;   //§&szlig;§é§à§é¨f§3

const int inf=0x3f3f3f3f;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//**************************************************************************************

struct node

{

    int x,y;

};

int dis[maxn][];

vector<node> E[maxn];

int inq[maxn][];

int main()

{

    int n=read(),m=read(),k=read(),s=read(),t=read();

    for(int i=;i<m;i++)

    {

        int x=read(),y=read(),z=read();

        E[x].push_back((node){y,z});

        E[y].push_back((node){x,z});

    }

    for(int i=;i<=n;i++)

        for(int j=;j<=k;j++)

            dis[i][j]=inf;

    queue<node> Q;

    Q.push((node){s,});

    dis[s][]=;

    int ans = inf;

    while(!Q.empty())

    {

        node now = Q.front();

        Q.pop();

        inq[now.x][now.y]=;

        for(int i=;i<E[now.x].size();i++)

        {

            if(dis[now.x][now.y]+E[now.x][i].y<dis[E[now.x][i].x][now.y])

            {

                dis[E[now.x][i].x][now.y]=dis[now.x][now.y]+E[now.x][i].y;

                if(!inq[E[now.x][i].x][now.y])

                {

                    inq[E[now.x][i].x][now.y]=;

                    Q.push((node){E[now.x][i].x,now.y});

                }

            }

            if(dis[now.x][now.y]<dis[E[now.x][i].x][now.y+]&&now.y<k)

            {

                dis[E[now.x][i].x][now.y+]=dis[now.x][now.y];

                if(!inq[E[now.x][i].x][now.y+])

                {

                    inq[E[now.x][i].x][now.y+] = ;

                    Q.push((node){E[now.x][i].x,now.y+});

                }

            }

        }

    }

    for(int i=;i<=k;i++)

        ans = min(ans,dis[t][i]);

    cout<<ans<<endl;

}