HDU3038 - How Many Answers Are Wrong(带权并查集)
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题目链接 HDU3038
【题意】给出M个[a,b]之间的和为c,如果和前面冲突则不处理,问一共有多少个冲突。
【分析】 感觉这种并查集比较难呀,想了好久还没完全明白。用sum[i]记录i到它的父亲节点的距离,然后用并查集检查输入的a,b点所在的集合,如果不在一个集合就合并,如果在一个集合说明已经出现过,检查sum[b]-sum[a](当前已知的[a,b]的和)是否等于输入的c,不等于则冲突;关键是找到如何更新sum[],对于不在一个集合的话,需要合并集合,对于sum[y] = sum[a]-sum[b]+c 这个公式是由向量得到的看下图就能明白了。对于find时更新,这个很普通,sum[x] += sum[f[x]]即可
【AC CODE】46ms
#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <queue>#include <stack>#include <vector>#include <string>#include <algorithm>using namespace std;#define rep(i,a,n) for(int i = a; i < n; i++)#define repe(i,a,n) for(int i = a; i <= n; i++)#define per(i,n,a) for(int i = n; i >= a; i--)#define clc(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3ftypedef long long LL;#define MAXN 50100int f[MAXN], sum[MAXN];int find(int x){if (f[x] == x) return x;int tmp = find(f[x]);sum[x] += sum[f[x]];return f[x] = tmp;}int main(){#ifdef SHYfreopen("e:\\1.txt", "r", stdin);#endifint n, m;while (~scanf("%d %d%*c", &n, &m)){int a, b, c, ans = 0;repe(i, 0, n) f[i] = i;clc(sum, 0);rep(i, 0, m){scanf("%d %d %d%*c", &a, &b, &c);int x = find(a), y = find(b);if (x != y){sum[y] = sum[a] + c - sum[b];f[y] = x;}else if (sum[b] - sum[a] != c)ans++;}printf("%d\n", ans);}return 0;}
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