HDU-3038.How Many Answers Are Wrong(带权并查集)

[3038.How Many Answers Are Wrong)(http://acm.hdu.edu.cn/showproblem.php?pid=3038)

题目大意

这题大概意思就是，通过前面所给区间的和，推算所给下一个区间的和的正确与否。

Input

Line 1: Two integers, N and M (1 <= N <= 200000, 1 <= M <= 40000). Means TT wrote N integers and FF asked her M questions.

Line 2…M+1: Line i+1 contains three integer: Ai, Bi and Si. Means TT answered FF that the sum from Ai to Bi is Si. It’s guaranteed that 0 < Ai <= Bi <= N.

You can assume that any sum of subsequence is fit in 32-bit integer.

Output

A single line with a integer denotes how many answers are wrong.

Sample Input

Sample Output

带权并查集

#include<cstdio>
#include<cstring>
const int maxn = 200000+5;
int fa[maxn];
int sum[maxn]; //表示根结点到子节点的和
int n,m; //n个数，m个问题
int find(int x)
{ 
    if(x!=fa[x])
    { 
        int root=find(fa[x]);
        sum[x]+=sum[fa[x]]; //权值合并
        fa[x]=root; //路径压缩
    }
    return fa[x];
}
void init() //初始化
{ 
    for(int i=1;i<=n;i++)
    { 
        fa[i] = i;
    }
}
int main()
{ 
    while(~scanf("%d%d",&n,&m))
    { 
        int ans=0;
        memset(sum, 0, sizeof(sum));
        init();
        for(int i=0;i<m;i++)
        { 
            int l,r,d; //l表示左边界，r表示右边界，d表示权值
            scanf("%d%d%d",&l,&r,&d);
            l--; //左闭右开区间
            int fl=find(l);
            int fr=find(r);
            if(fl!=fr) //该区间未能通过前面所给区间推出来
            { 
                fa[fr]=fl;
                sum[fr]=sum[l]-sum[r]+d;
            }
            else if(sum[r]-sum[l]!=d) 
            { 
                ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}