• 地铁 Dijkstra(优先队列优化) 湖南省第12届省赛


    传送门:地铁

    思路:拆点,最短路;拆点比较复杂,所以对边进行最短路,spfa会tle,所以改用Dijkstra(优先队列优化)  

    模板

    /**************************************************************
        Problem:
        User: youmi
        Language: C++
        Result: Accepted
        Time:
        Memory:
    ****************************************************************/
    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    //#include<bits/stdc++.h>
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <stack>
    #include <set>
    #include <sstream>
    #include <cmath>
    #include <queue>
    #include <deque>
    #include <string>
    #include <vector>
    #define zeros(a) memset(a,0,sizeof(a))
    #define ones(a) memset(a,-1,sizeof(a))
    #define sc(a) scanf("%d",&a)
    #define sc2(a,b) scanf("%d%d",&a,&b)
    #define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
    #define scs(a) scanf("%s",a)
    #define sclld(a) scanf("%I64d",&a)
    #define pt(a) printf("%d
    ",a)
    #define ptlld(a) printf("%I64d
    ",a)
    #define rep(i,from,to) for(int i=from;i<=to;i++)
    #define irep(i,to,from) for(int i=to;i>=from;i--)
    #define Max(a,b) ((a)>(b)?(a):(b))
    #define Min(a,b) ((a)<(b)?(a):(b))
    #define lson (step<<1)
    #define rson (lson+1)
    #define eps 1e-6
    #define oo (1e16)
    #define TEST cout<<"*************************"<<endl
    const double pi=4*atan(1.0);
    
    using namespace std;
    typedef long long ll;
    template <class T> inline void read(T &n)
    {
        char c; int flag = 1;
        for (c = getchar(); !(c >= '0' && c <= '9' || c == '-'); c = getchar()); if (c == '-') flag = -1, n = 0; else n = c - '0';
        for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0'; n *= flag;
    }
    ll Pow(ll base, ll n, ll mo)
    {
        if (n == 0) return 1;
        if (n == 1) return base % mo;
        ll tmp = Pow(base, n >> 1, mo);
        tmp = (ll)tmp * tmp % mo;
        if (n & 1) tmp = (ll)tmp * base % mo;
        return tmp;
    }
    //***************************
    
    int n,m;
    const int maxn=100000+10;
    const ll mod=1000000007;
    struct node
    {
        int v,c;
        ll t,dis;
        node(){}
        node(int _v,int _c,ll _t,ll _dis)
        {
            v=_v,c=_c,t=_t,dis=_dis;
        }
        bool operator<(const node &a)const
        {
            if(dis==a.dis)
                return t>a.t;
            return dis>a.dis;
        }
    };
    vector<node>vt[maxn];
    void dijkstra()
    {
        ll ans=oo;
        priority_queue<node>q;
        for(int i=0;i<vt[1].size();i++)
        {
            vt[1][i].dis=vt[1][i].t;
            if(vt[1][i].v==n)
                ans=min(ans,vt[1][i].dis);
            q.push(vt[1][i]);
        }
        while(!q.empty())
        {
            node cur=q.top();
            q.pop();
            for(int i=0;i<vt[cur.v].size();i++)
            {
                node temp=vt[cur.v][i];
                if(temp.dis>cur.dis+temp.t+abs(cur.c-temp.c))
                {
                    temp.dis=cur.dis+temp.t+abs(cur.c-temp.c);
                    if(temp.v==n)
                        ans=min(ans,temp.dis);
                    vt[cur.v][i]=temp;//非常重要,不要忘了这一步
                    q.push(temp);
                }
            }
        }
        printf("%lld
    ",ans);
    }
    int main()
    {
        #ifndef ONLINE_JUDGE
        //freopen("in.txt","r",stdin);
        #endif
        while(~sc2(n,m))
        {
            int a,b,c;
            ll t;
            for(int i=1;i<=n;i++)
                vt[i].clear();
            rep(i,1,m)
            {
                sc3(a,b,c);
                scanf("%lld",&t);
                vt[a].push_back(node(b,c,t,oo));
                vt[b].push_back(node(a,c,t,oo));
            }
            dijkstra();
        }
    }
    不为失败找借口,只为成功找方法
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  • 原文地址:https://www.cnblogs.com/youmi/p/5860491.html
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