hdu 4496 并查集 逆向 并查集删边

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

 1 #include 
 2 #include 
 3 #include 
 4 
 5 const int maxn = 10000 + 500;
 6 const int maxm = 100000 + 5000;
 7 int father[maxn];
 8 int x1[maxm], x2[maxm];
 9 int ans[maxm];
10 int n, m;
11 
12 
13 int getfather(int x) {
14     if (father[x] == x) return (x);
15     return (father[x] = getfather(father[x]));
16 }
17 
18 int main () {
19     while(scanf("%d %d", &n, &m) != EOF) {
20         memset(ans, 0, sizeof(ans));
21         for (int i = 1; i <= n; i++) father[i] = i;
22         for (int i = 1; i <= m; i++) {
23             scanf("%d %d", &x1[i], &x2[i]);
24             x1[i] += 1;
25             x2[i] += 1;
26         }
27         ans[m] = n;
28         for (int i = m; i >= 1; i--) {
29             int tx = getfather(x1[i]);
30             int ty = getfather(x2[i]);
31             if (tx != ty) {
32                 ans[i-1] = ans[i] - 1;
33                 father[tx] = ty;
34             } else {
35                 ans[i-1] = ans[i];
36             }
37         }
38         for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
39     }
40     return 0;
41 }