bzoj4690 Never Wait for Weights

Posted 2020-09-20 逢山开路 遇水架桥

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4690

【题解】

带权并查集

fa[x]表示x的父亲，a[x]表示x到x的父亲多/少多少

那么找祖先的时候算一下到祖先多少，然后路径压缩。

合并的时候注意让fa[fx]=fy的时候，a[fx]是多少（这个可以算的）

然后查询直接做就行了。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, q;

int fa[M], a[M]; 
inline int getf(int x) {
    int i=x, j, jj, res = 0;
    while(fa[x] != x) {
        res = res + a[x];
        x = fa[x];
    }
    while(i != x) {
         j = fa[i]; jj = a[i]; 
        fa[i] = x;
        a[i] = res;
        res -= jj;
        i = j;
    }
    return x;
}

int main() {
    while(cin >> n >> q && (n+q)) { 
        for (int i=1; i<=n; ++i) fa[i] = i, a[i] = 0;
        char opt[3]; int x, y, z; 
        while(q--) {
            scanf("%s", opt);
            if(opt[0] == ‘!‘) {
                scanf("%d%d%d", &x, &y, &z);
                int fx = getf(x), fy = getf(y);
                if(fx == fy) continue;
                fa[fx] = fy;
                a[fx] = z-a[x]+a[y]; 
            } else {
                scanf("%d%d", &x, &y);
                int fx = getf(x), fy = getf(y); 
                if(fx != fy) puts("UNKNOWN");
                else printf("%d\n", a[x]-a[y]); 
            }
        }
    }

    return 0;
}

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