貌似某大犇说过 正难则反,,, 题目说要对这张图进行删边,然后判断联通块的个数,那么就可以先把所有边都删掉,之后从后往前加边,若加的边两端点不在同一个联通块中, 那么此时联通快个数少一,否则不变 #include <cstdio> #include <cstring> #include <algorithm> + ; + ; int father[maxn]; int x1[maxm], x2[maxm]; int ans[maxm]; int n, m; int ge…
Stability Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 65535/102400 K (Java/Others)Total Submission(s): 1347 Accepted Submission(s): 319 Problem Description Given an undirected connected graph G with n nodes and m edges, with possibly r…