每日一題 — 2020 - 05 - 06

今天的每日一題確實有點坑，可能評測姬炸了，不過又學到最短路了

題目鏈接

推理過程：

• 這個題有點小思維，其實也沒啥，就是從起始點到每個點的距離，用白天的費用算一遍，然後用終止點到每個點的距離的費用，用黑夜的費用算一遍
• 然後由於有個限制條件，所以我們找最小值的時候應該從後往前找。
• 交了差不多30發，不過學到了unordered_map 以後查詢就用他了

代碼：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
#include <unordered_map>
using namespace std;
   
typedef pair<int,int> PII;
const int N = 100010, M = 200010;
   
int n ,m,idx;
   
int e[M],ne[M],h[N];
   
long long dist1[M], dist2[M], res[N];
   
bool st[N];
   
unordered_map<int,map<int,long long> >mp1, mp2;
   
void add(int x,int y){
    e[idx] = y, ne[idx] = h[x],h[x] = idx++;
}
   
void spfa1(int u){
    memset(dist1,0x3f,sizeof dist1);
    dist1[u] = 0;
    queue <int> q;
    q.push(u);
    st[u] = true;
    while(q.size()){
        int t = q.front();
        q.pop();
        st[t] = false;
        for (int i = h[t]; i != -1; i = ne[i]){
            int j = e[i];
            if (dist1[j] > dist1[t] + mp1[t][j]){
                dist1[j] = dist1[t] + mp1[t][j];
                if (!st[j]){
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
       
}
   
void spfa2(int u){
    memset(dist2,0x3f,sizeof dist2);
    dist2[u] = 0;
    queue <int> q;
    q.push(u);
    st[u] = true;
    while(q.size()){
        int t = q.front();
        q.pop();
        st[t] = false;
        for (int i = h[t]; i != -1; i = ne[i]){
            int j = e[i];
            if (dist2[j] > dist2[t] + mp2[t][j]){
                dist2[j] = dist2[t] + mp2[t][j];
                if (!st[j]){
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
       
}
   
int main(){
    scanf("%d%d",&n,&m);
    memset(h,-1,sizeof h);
    for (int i = 0; i < m; i ++){
        int x , y , a, b;
        scanf("%d%d%d%d",&x,&y,&a,&b);
        add(x,y);
        add(y,x);
        mp1[x][y] = a;
        mp1[y][x] = a;
        mp2[x][y] = b;
        mp2[y][x] = b;
    }
    int t, s;
    scanf("%d%d",&t,&s);
    spfa1(t);
    memset(st,false,sizeof st);
    spfa2(s);
       
   
    long long mx = 1e18;
   
    for (int i = n; i >= 1; i --){
        mx = min(mx,dist1[i] + dist2[i]);
        res[i] = mx;
    }
   
    for (int i = 1; i <= n; i ++){
        printf("%lld\n",res[i]);
    }
    return 0;
}