带权并查集
设cnt[x]为x到本集合代表元素的路径值,在递归寻找代表元素时，如果找到就直接返回，没找到先记录下此时x的fa[x]，再让fa[x]被更新，fa[x]被更新也意味着fa[x]到代表元素的cnt[fa[x]]被更新了，则此时再用cnt[fa[x]]更新cnt[x]，便完成了路径更新。
核心代码
~cpp~
inline int f(int x) {
if(x ^ fa[x]) {
int t = fa[x];
fa[x] = f(fa[x]);
cnt[x] += cnt[t];
}
return fa[x];
}
How Many Answers Are Wrong HDU - 3038
题意:给出一个区间的长度 N，及 M 个子区间和， 形如：x y z, 表示子区间 [x, y] 的和为 z
如果一个“子区间和”与前面的“子区间和”冲突，即为错误（而且这个“子区间和”将在接下来的判断中被忽略）。
求总错误个数。
~~~
include
include
include
include
include
include
include
define rint register int
define ll long long
define lson x<<1
define rson x<<1|1
define mid ((st[x].l + st[x].r) >> 1)
using namespace std;
template
{
int f = 1;x = 0;
char c = getchar();
for(; c < '0' || c > '9' ; c = getchar()) if(c=='-') f = -1;
for(;'0' <= c && c <= '9'; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
x = f;
}
template
{
if(x < 0) {
putchar('-');
x = -x;
}
if(x > 9) print(x/10);
putchar(x % 10 + '0');
}
const int inf = 0x7fffffff;
const int maxn = 200100;
const int mod = 1000000007;
int n,m;
int fa[maxn];
int cnt[maxn];//自身到集合代表的权值
inline int f(int x) {
if(x ^ fa[x]) {
int t = fa[x];
fa[x] = f(fa[x]);
cnt[x] += cnt[t];
}
return fa[x];
}
int main()
{
while(~scanf("%d%d",&n,&m)) {
for(rint i = 1;i <= n; ++i) {
fa[i] = i;
cnt[i] = 0;
}
int ans = 0;
for(rint i = 1;i <= m; ++i) {
int x,y,z;
read(x);read(y);read(z);
--x;
int fx = f(x);
int fy = f(y);
if(fx == fy) {
if(cnt[x] - cnt[y] != z) ++ans;
}
else {
fa[fx] = fy;
cnt[fx] = cnt[y] - cnt[x] + z;
}
}
print(ans);putchar('\n');
}
}
/
*/
~~~