• 最短路径(步行于乘车)


    http://codeforces.com/gym/101061/problem/C

    题意:有n个十字路口 , m条双向道路,步行道和车道,问从u到v,要使步行路程尽可能的小,如果步行路程相同,则总路程尽可能小。

    解法:以步行为第一优先级,车程为第二优先级,有车道乘车则步行就赋值为0,如果有步行无车道则该车道赋值为0。

    //#include<bits/stdc++.h>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <stdio.h>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <string.h>
    #include <vector>
    #include <stdlib.h>
    using namespace std;
    typedef long long ll ;
    #define int ll
    #define mod 100
    #define gcd(m,n) __gcd(m, n)
    #define rep(i , j , n) for(int i = j ; i <= n ; i++)
    #define red(i , n , j)  for(int i = n ; i >= j ; i--)
    #define ME(x , y) memset(x , y , sizeof(x))
    //int lcm(int a , int b){return a*b/gcd(a,b);}
    //ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
    //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
    //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
    #define INF  0x3f3f3f3f
    #define PI acos(-1)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define lson l,mid,root<<1
    #define rson mid+1,r,root<<1|1
    #define pb push_back
    #define mp make_pair
    #define all(v) v.begin(),v.end()
    #define size(v) (int)(v.size())
    #define cin(x) scanf("%lld" , &x);
    const int N = 1e5+9;
    const int maxn = 1e2+9;
    const double esp = 1e-6;
    int n , m ;
    int c[maxn][maxn] , p[maxn][maxn] , vis[maxn] , dis[maxn];
    int val[maxn];
    
    void dijkstra(int u){
        rep(i , 1 , n){
            dis[i] = p[u][i];
            val[i] = c[u][i];
        }
        vis[u] = 1 ;
        rep(i , 1 , n-1){
            int pos = -1;
            int map = INF , mac = INF;
            rep(j , 1 , n){
                if(!vis[j] && map > dis[j]){
                    map = dis[j];
                    mac = val[j];
                    pos = j ;
                }else if(!vis[j] && map == dis[j] && mac > val[j]){
                    mac = val[j];
                    pos = j ;
                }
            }
            if(pos == -1) break;
            vis[pos] = 1 ;
            rep(j , 1 , n){
                if(!vis[j] && dis[j] > dis[pos] + p[pos][j]){
                    dis[j] = dis[pos] + p[pos][j];
                    val[j] = val[pos] + c[pos][j];
                }else if(!vis[j] && dis[j] == dis[pos] + p[pos][j] && val[j] > val[pos] + c[pos][j]){
                    val[j] = val[pos] + c[pos][j];
                }
            }
        }
    }
    
    void init(){
        ME(vis , 0);
        fill(p[0] , p[0]+maxn*maxn , INF);
        fill(c[0] , c[0]+maxn*maxn , INF);
    }
    void solve(){
        init();
        scanf("%lld%lld" , &n , &m);
        rep(i , 1 , m){
            int u , v , w , k ;
            scanf("%lld%lld%lld%lld" , &u , &v , &w ,&k);
            if(k == 1){
                p[u][v] = p[v][u] = min(p[u][v] , w);
            }else{
                c[u][v] = c[v][u] = min(c[u][v] , w);
            }
        }
        rep(i , 1 , n){
            rep(j , 1 , n){
                if(c[i][j] != INF){
                    p[i][j] = 0 ;
                }else if(c[i][j] == INF && p[i][j] != INF){
                    c[i][j] = 0 ;
                }
            }
        }
        int u , v ;
        scanf("%lld%lld" , &u , &v);
        dijkstra(u);
        if(dis[v] == INF && val[v] == INF) cout << -1 << endl;
        else cout << dis[v] << " " << dis[v]+val[v] << endl;
    }
    
    signed main()
    {
        int t ;
        cin >> t ;
        while(t--){
            solve();
        }
    }
    
  • 相关阅读:
    版本控制之GitHub亲手实验总结
    Java的HashMap是如何实现的?
    Junit
    由swap引发的关于按值传递和引用传递的思考与总结
    C++了解free和delete
    GitHub使用教程
    Oracle下SQL学习笔记
    Flappy Bird
    尾递归与Continuation(转载)
    十步完全理解SQL(转载)
  • 原文地址:https://www.cnblogs.com/nonames/p/11379630.html
Copyright © 2020-2023  润新知