guojiaqi007的专栏

http://ac.jobdu.com/problem.php?pid=1100 参考了http://blog.youkuaiyun.com/thudaliangrx/article/details/49357075 因为每条新边比之前所有边的和都要长，所以只要是当前边能联通的点对，点对当前的距离就是最短距离 #include static int folk[102]; st

https://www.patest.cn/contests/pat-a-practise/1111 #include #include #include #include #include using namespace std; int timeC[500][500],len[500][500],visit[500],dist[500], from[500],nNode[500

https://www.patest.cn/contests/pat-a-practise/1003 #include #include #include using namespace std; pair dijstra(vector> g, vector hands,int c1,int c2,int n) { vector v(n, 0); vector dist(n, IN

http://ac.jobdu.com/problem.php?pid=1008 这道题告诉我们 INT_MAX, 慎用，dijkstra 有时加溢出了，你都不知道 #include //#include #include using namespace std; #define rep(i,j,k) for(int i=j;i<=k;i++) #define inone(i)