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hdu 4496 并查集 逆向 并查集删边

2016-11-08 11:19 176 查看
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 <cstdio>
2 #include <cstring>
3 #include <algorithm>
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 }


 
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