hdu 3038 How Many Answers Are Wrong (种类并查集)
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解题思路:这是并查集里面很特殊的一类题目,不仅仅需要考虑点与点之间的集合关系,点与点之间还有一定的关系,姑且把他称之为边权,解决这个问题就是要用到这种思想,针对每一条线段,我们把两个端点提取出来,把他们的和作为边权,然后建立并查集,如果一个线段的两个端点已经在同一个根节点下,那么根据定义是可以得到这条线段的值的,如果不在一个根节点,可以将两个集合合并,然后维护两个根节点形成的新的线段。
AC代码:
/* @Author: wchhlbt @Date: 2017/4/28*/#include <bits/stdc++.h>#define Fori(x) for(int i=0;i<x;i++)#define Forj(x) for(int j=0;j<x;j++)#define maxn 200007#define inf 0x3f3f3f3f#define ONES(x) __builtin_popcount(x)using namespace std;typedef long long ll ;const double eps =1e-8;const int mod = 1000000007;typedef pair<int, int> P;const double PI = acos(-1.0);int dx[4] = {0,0,1,-1};int dy[4] = {1,-1,0,0};int n,m;int ans;int f[maxn];//保存每个线段的起点int sum[maxn];//保存当前节点到根节点的路径长度,每条线段的和存放在子节点上void init()//初始化{ for(int i = 0; i<maxn-2; i++) f[i] = i , sum[i] = 0;}int find(int u)//更新父节点{ if(u!=f[u]){ int t = f[u]; f[u] = find(f[u]); sum[u] += sum[t];//同时维护子节点到根节点的路径长度(从根节点到它的线段和) } return f[u];}void connect(int l, int r, int s){ int x = find(l); int y = find(r); if(x!=y){ f[x] = y; sum[x] = sum[r] - sum[l] - s;//确定新建立联系的两个节点的路径长度 } else{ if(sum[r]-sum[l]!=s)//确定冲突条件 ans++; }}int main(){ //freopen("test.txt","r",stdin); while(~scanf("%d%d",&n,&m)) { ans = 0; init(); int l,r,s; for(int i = 0; i<m; i++){ scanf("%d%d%d",&l,&r,&s); l--;//将闭区间转化为左开右闭,保证区间有交集 connect(l,r,s); } printf("%d\n",ans); } return 0;}/*unsigned int 0~4294967295int 2147483648~2147483647unsigned long 0~4294967295long 2147483648~2147483647long long的最大值:9223372036854775807long long的最小值:-9223372036854775808unsigned long long的最大值:18446744073709551615__int64的最大值:9223372036854775807__int64的最小值:-9223372036854775808unsigned __int64的最大值:18446744073709551615*/
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