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

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 const int maxn =  + ;
 const int maxm =  + ;
 int father[maxn];
 int x1[maxm], x2[maxm];
 int ans[maxm];
 int n, m;

 int getfather(int x) {
     if (father[x] == x) return (x);
     return (father[x] = getfather(father[x]));
 }

 int main () {
     while(scanf("%d %d", &n, &m) != EOF) {
         memset(ans, , sizeof(ans));
         for (int i = ; i <= n; i++) father[i] = i;
         for (int i = ; i <= m; i++) {
             scanf("%d %d", &x1[i], &x2[i]);
             x1[i] += ;
             x2[i] += ;
         }
         ans[m] = n;
         for (int i = m; i >= ; i--) {
             int tx = getfather(x1[i]);
             int ty = getfather(x2[i]);
             if (tx != ty) {
                 ans[i-] = ans[i] - ;
                 father[tx] = ty;
             } else {
                 ans[i-] = ans[i];
             }
         }
         for (int i = ; i <= m; i++) printf("%d\n", ans[i]);
     }
     return ;
 }