【[Offer收割]编程练习赛15 B】分数调查

【题目链接】:http://hihocoder.com/problemset/problem/1515

【题意】

【题解】



带权并查集

relation[x]表示父亲节点比当前节点大多少;

对于输入的x,y,z;

如果z小于0;

则交换x,y同时z取相反数;

然后按照带权并查集的更新方式,对y的根节点的r2的relation[r2]进行更新即可;



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,0,-1,0,-1,-1,1,1};

const int dy[9] = {0,0,-1,0,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 100000+100;

const int INF = 0x3f3f3f3f;

int n,m,q,f[N],relation[N];

int ff(int x)

{

    if (f[x] == x)

        return x;

    int olfa = f[x];

    f[x] = ff(f[x]);

    relation[x] = relation[x] + relation[olfa];

    return f[x];

}

int main()

{

    //freopen("D:\\rush.txt","r",stdin);

    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use

    cin >> n >> m >> q;

    rep1(i,1,n)

        f[i] = i,relation[i] = 0;

    rep1(i,1,m)

    {

        int x,y,z;

        cin >> x >> y >> z;

        if (z<0)

        {

            swap(x,y);

            z = -z;

        }

        int r1 = ff(x),r2 = ff(y);

        if (r1!=r2)

        {

            f[r2] = r1;

            relation[r2] = z+relation[x]-relation[y];

        }

    }

    rep1(i,1,q)

    {

        int x,y;

        cin >> x >> y;

        int r1 = ff(x),r2 = ff(y);

        if (r1!=r2)

            cout << -1 << endl;

        else

        {

            int temp = relation[y]-relation[x];

            cout << temp << endl;

        }

    }

    return 0;

}