As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are…

最短路+dfs 先找出可能在最短路上的边,这些边会构成一个DAG,然后在这个DAG上dfs一次就可以得到两个答案了. 也可以对DAG进行拓扑排序,然后DP求解. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<queue> #include<vector> using nam…

1003. Emergency (25) As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between an…

Emergency As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of c…

二次探测法.表示第一次听说这东西... #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; ; bool fl…

简单贪心.先买性价比高的. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; ; +; int n; str…

scanf读入居然会超时...用了一下输入挂才AC... #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<queue> #include<iostream> #include<algorithm> using namespace std; +; int a[maxn],b[maxn]; int n1,n2; inlin…

撸完这题,感觉被掏空. 由于进制可能大的飞起..所以需要开longlong存,答案可以二分得到. 进制很大,导致转换成10进制的时候可能爆long long,在二分的时候,如果溢出了,那么上界=mid-1 #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> using namespace std; ],t[]; int…

简单题,不过数据中好像存在有环的链表...... #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<vector> using namespace std; +; struct Node { int id;…

题意:n个点,m条双向边,每条边给出通过用时,每个点给出点上的人数,给出起点终点,求不同的最短路的数量以及最短路上最多能通过多少人.(N<=500) AAAAAccepted code: #include<bits/stdc++.h> using namespace std; int n,m; int s,t; ]; vector<pair<]; vector<]; vector<int>tmp_path; ; ]; ]; ; void SPFA(){ qu…