吉林省赛及东北四省赛 模板

目录

• 线段树
  • 区间和，单点修改
  • 区间和，区间修改
  • 区间乘加
  • 区间最大值，无修改
• 三分（凸函数）
• 快速幂
• 矩阵快速幂
• KMP
• 求逆元
  • 模为素数
  • 模不为素数
  • 求一段逆元
• LCA
  • tarjan离线算法
• FFT
• Java大数
• 去重离散化
  • 浮点数
  • 整数
• Cantor展开
• spfa, $O(kE)/O(VE)$
• dijkstra
  • 一般实现，$O(V^2)$
  • 堆优化实现，$O(ElogV)$
• 带权并查集
• 尾声

线段树

区间和，单点修改

#include <bits/stdc++.h>
typedef long long LL;
const int maxn = 500000;

using namespace std;

struct tTree
{
    LL sum;
    int l, r;
};
tTree tree[maxn*4+10];

void pushup(int x)
{
    if(tree[x].l == tree[x].r)
        return;
    tree[x].sum = tree[x*2].sum +tree[x*2+1].sum;
}

void build(int x, int l, int r)
{
    tree[x].l = l; tree[x].r = r;
    if(tree[x].l == tree[x].r)
        scanf("%lld", &tree[x].sum);
    else
    {
        int mid = (l+r)/2;
        build(x*2, l, mid);
        build(x*2+1, mid+1, r);
        pushup(x);
    }
}

void modify(int x, int pos, int val)
{
    if(tree[x].l == tree[x].r)
        tree[x].sum += val;
    else
    {
        int mid = (tree[x].l+tree[x].r)/2;
        if(pos <= mid)
            modify(x*2, pos, val);
        else
            modify(x*2+1, pos, val);
        pushup(x);
    }
}

LL query(int x, int l, int r)
{
    if(tree[x].l>=l && tree[x].r<=r)
        return tree[x].sum;
    else
    {
        int mid = (tree[x].l+tree[x].r)/2;
        LL ret = 0;
        if(mid >= l)
            ret += query(x*2, l, r);
        if(mid+1 <= r)
            ret += query(x*2+1, l, r);
        return ret;
    }
}

int main()
{
    int N, M;
    scanf("%d%d", &N, &M);
    build(1, 1, N);
    for(int i=1,op,x,y; i<=M; i++)
    {
        scanf("%d%d%d", &op, &x, &y);
        if(op == 1)
        {
            modify(1, x, y);
        }
        if(op == 2)
        {
            printf("%lld
", query(1, x, y));
        }
    }
    return 0;
}

区间和，区间修改

#include <bits/stdc++.h>
typedef long long LL;
const int maxn = 500000;

using namespace std;

struct tTree
{
    LL sum, tag;
    int l, r, len;
};
tTree tree[maxn*4+10];

void pushup(int x)
{
    if (tree[x].l == tree[x].r)
        return;
    tree[x].sum = tree[x*2].sum + tree[x*2+1].sum;
}

void pushdown(int x)
{
    if (tree[x].l == tree[x].r)
        return;
    LL tag = tree[x].tag;
    tree[x].tag = 0;
    tree[x*2].sum += tag*tree[x*2].len;
    tree[x*2+1].sum += tag*tree[x*2+1].len;
    tree[x*2].tag += tag;
    tree[x*2+1].tag += tag;
}

void build(int x, int l, int r)
{
    tree[x].l = l; tree[x].r = r;
    tree[x].tag = 0;
    tree[x].len = r-l+1;
    if (tree[x].l == tree[x].r)
        scanf("%lld", &tree[x].sum);
    else
    {
        int mid = (l+r)/2;
        build(x*2, l, mid);
        build(x*2+1, mid+1, r);
        pushup(x);
    }
}

void modify(int x, int l, int r, int val)
{
    if (tree[x].l>=l && tree[x].r<=r)
    {
        tree[x].tag += val;
        tree[x].sum += val*tree[x].len;
    }
    else
    {
        pushdown(x);
        int mid = (tree[x].l+tree[x].r)/2;
        if (mid >= l)
            modify(x*2, l, r, val);
        if (mid+1 <= r)
            modify(x*2+1, l, r, val);
        pushup(x);
    }
}

//the sum of the intersection of tree[x].l-tree[x].r and l-r
LL query(int x, int l, int r)
{
    if (tree[x].l>=l && tree[x].r<=r)
        return tree[x].sum;
    else
    {
        pushdown(x);
        int mid = (tree[x].l+tree[x].r)/2;
        LL ret = 0;
        if (mid >= l)
            ret += query(x*2, l, r);
        if (mid+1 <= r)
            ret += query(x*2+1, l, r);
        return ret;
    }
}

int main()
{
    int N, M;
    scanf("%d%d", &N, &M);
    build(1, 1, N);
    for (int i=1,op,x,y,k; i<=M; i++)
    {
        scanf("%d", &op);
        if (op == 1)
        {
            scanf("%d%d%d", &x, &y, &k);
            modify(1, x, y, k);
        }
        if (op == 2)
        {
            scanf("%d", &x);
            printf("%lld
", query(1, x, x));
        }
    }
    return 0;
}

区间乘加

#include <bits/stdc++.h>
typedef long long LL;
const int maxn = 100000;

using namespace std;

int N, M; LL P;

struct tTree
{
    LL sum, mul, add;
    int l, r, len;
};
tTree tree[maxn*4+10];

void pushup(int x)
{
    if (tree[x].l == tree[x].r)
        return;
    tree[x].sum = (tree[x*2].sum + tree[x*2+1].sum)%P;
}

//changes of sum, mul, add
void pushdown(int x)
{
    if (tree[x].l == tree[x].r)
        return;
    LL mul = tree[x].mul;
    LL add = tree[x].add;
    tree[x].mul = 1;
    tree[x].add = 0;
    tree[x*2].sum = (tree[x*2].sum*mul%P + add*tree[x*2].len%P)%P;
    tree[x*2+1].sum = (tree[x*2+1].sum*mul%P + add*tree[x*2+1].len%P)%P;
    tree[x*2].mul = tree[x*2].mul*mul%P;
    tree[x*2].add = (tree[x*2].add*mul%P + add)%P;
    tree[x*2+1].mul = tree[x*2+1].mul*mul%P;
    tree[x*2+1].add = (tree[x*2+1].add*mul%P + add)%P;
}

void build(int x, int l, int r)
{
    tree[x].l = l; tree[x].r = r;
    tree[x].len = r-l+1;
    tree[x].mul = 1; tree[x].add = 0;
    if (l == r)
        scanf("%lld", &tree[x].sum);
    else
    {
        int mid = (l+r)/2;
        build(x*2, l, mid);
        build(x*2+1, mid+1, r);
        pushup(x);
    }
}

void modify(int x, int l, int r, int op, int val)
{
    //keep the sum right
    if (tree[x].l>=l && tree[x].r<=r)
    {
        if (op == 1)
        {
            tree[x].sum = tree[x].sum*val%P;
            tree[x].add = tree[x].add*val%P;
            tree[x].mul = tree[x].mul*val%P;
        }
        if (op == 2)
        {
            tree[x].sum = (tree[x].sum+tree[x].len*val)%P;
            tree[x].add = (tree[x].add+val)%P;
        }
    }
    else
    {
        pushdown(x);
        int mid = (tree[x].l+tree[x].r)/2;
        if (l <= mid)
            modify(x*2, l, r, op, val);
        if (mid+1 <= r)
            modify(x*2+1, l, r, op, val);
        pushup(x);
    }
}

LL query(int x, int l, int r)
{
    if (tree[x].l>=l && tree[x].r<=r)
    {
        return tree[x].sum%P;
    }
    else
    {
        pushdown(x);
        int mid = (tree[x].l+tree[x].r)/2;
        LL ret = 0;
        if (l <= mid)
            ret = (ret+query(x*2, l, r))%P;
        if (mid+1 <= r)
            ret = (ret+query(x*2+1, l, r))%P;
        return ret%P;
    }
}

int main()
{
    scanf("%d%d%lld", &N, &M, &P);
    build(1, 1, N);
    for (int i=1,op,x,y,k; i<=M; i++)
    {
        scanf("%d", &op);
        if (op == 1)
        {
            scanf("%d%d%d", &x, &y, &k);
            modify(1, x, y, 1, k);
        }
        if (op == 2)
        {
            scanf("%d%d%d", &x, &y, &k);
            modify(1, x, y, 2, k);
        }
        if (op == 3)
        {
            scanf("%d%d", &x, &y);
            printf("%lld
", query(1, x, y));
        }
    }
    return 0;
}

区间最大值，无修改

#include <bits/stdc++.h>
const int maxn = 100000;

using namespace std;

struct tTree
{
    int mmax;
    int l, r;
};
tTree tree[maxn*4+10];

void pushup(int x)
{
    if (tree[x].l == tree[x].r)
        return;
    tree[x].mmax = max(tree[x*2].mmax, tree[x*2+1].mmax);
}

void build(int x, int l, int r)
{
    tree[x].l = l; tree[x].r = r;
    if (tree[x].l == tree[x].r)
        scanf("%d", &tree[x].mmax);
    else
    {
        int mid = (l+r)/2;
        build(x*2, l, mid);
        build(x*2+1, mid+1, r);
        pushup(x);
    }
}

int query(int x, int l, int r)
{
    if (tree[x].l>=l && tree[x].r<=r)
        return tree[x].mmax;
    else
    {
        int mid = (tree[x].l+tree[x].r)/2;
        int ret = -1;
        if (mid >= l)
            ret = max(ret, query(x*2, l, r));
        if (mid+1 <= r)
            ret = max(ret, query(x*2+1, l, r));
        return ret;
    }
}

int main()
{
    int N, M;
    scanf("%d%d", &N, &M);
    build(1, 1, N);
    for (int i=1,x,y; i<=M; i++)
    {
        scanf("%d%d", &x, &y);
        printf("%d
",query(1, x, y));
    }
    return 0;
}

三分（凸函数）

#include <bits/stdc++.h>
const double eps = 1e-7;

using namespace std;

int N; double l, r;
double a[20];

double f(double x)
{
    double ret = 0;
    double tempx = 1;
    for(int i=N+1; i>=1; i--)
    {
        ret += tempx*a[i];
        tempx *= x;
    }
    return ret;
}

double triSection(double l, double r)
{
    while (r-l > eps)
    {
        double mid1 = (l*2+r)/3;
        double mid2 = (l+r*2)/3;
        if (f(mid1) < f(mid2))
            l = mid1;
        else
            r = mid2;
    }
    return l;
}

int main()
{
    scanf("%d%lf%lf", &N, &l, &r);
    for(int i=1; i<=N+1; i++)
        scanf("%lf", a+i);
    printf("%.5f
", triSection(l,r));
    return 0;
}

快速幂

LL quickPow(LL a, LL b)
{
    LL ans = 1, base = a;
    while (b != 0)
    {
        if ((b&1) != 0)
            ans = (ans*base) % mod;
        base = (base*base) % mod;
        b >>= 1;
    }
    return ans;
}

矩阵快速幂

#include <bits/stdc++.h>
typedef long long LL;
const int maxn = 100;
const LL mod = 1000000007;

using namespace std;

int n; LL k;

struct tMatrix
{
    LL mat[maxn+10][maxn+10];
    tMatrix()
    {
        memset(mat, 0, sizeof(mat));
    }
    void build()
    {
        for (int i=1; i<=n; i++)
            mat[i][i] = 1;
    }
};

tMatrix operator*(const tMatrix& a, const tMatrix& b)
{
    tMatrix c;
    for (int k=1; k<=n; k++)
    for (int i=1; i<=n; i++)
    for (int j=1; j<=n; j++)
    {
        c.mat[i][j] = (c.mat[i][j]+a.mat[i][k]*b.mat[k][j]%mod)%mod;
    }
    return c;
}

tMatrix quickPowMat(tMatrix a, LL b)
{
    tMatrix ans; ans.build();
    tMatrix base = a;
    while (b != 0)
    {
        if ((b&1) != 0)
            ans = ans * base;
        base = base * base;
        b >>= 1;
    }
    return ans;
}

int main()
{
    scanf("%d%lld", &n, &k);
    tMatrix a;
    for (int i=1; i<=n; i++)
    for (int j=1; j<=n; j++)
    {
        scanf("%lld", &a.mat[i][j]);
    }
    tMatrix ans = quickPowMat(a, k);
    for (int i=1; i<=n; i++)
    for (int j=1; j<=n; j++)
    {
        printf("%lld", ans.mat[i][j]);
        if (j == n)
            printf("
");
        else
            printf(" ");
    }
    return 0;
}

KMP

#include <bits/stdc++.h>
const int maxn = 1000000;

using namespace std;

char a[maxn+10], b[maxn+10];
int la, lb;
int nxt[maxn+10];

void getNext()
{
    for (int i=2, j=0; i<=lb; i++)
    {
        while (j && b[j+1]!=b[i])
            j = nxt[j];
        if (b[j+1] == b[i])
            j++;
        nxt[i] = j;
    }
}

void KMP()
{
    //j means b[1..j] are successfully matched
    for (int i=1, j=0; i<=la; i++)
    {
        while (j && b[j+1]!=a[i])
            j = nxt[j];
        if (b[j+1] == a[i])
            j++;
        if (j == lb)
        {
            printf("%d
", i-lb+1);
            j = nxt[j];
        }
    }
}

int main()
{
    scanf("%s%s", a+1, b+1);
    la = strlen(a+1);
    lb = strlen(b+1);
    getNext();
    KMP();
    for (int i=1; i<=lb; i++)
    {
        if (i==lb)
            printf("%d
", nxt[i]);
        else
            printf("%d ", nxt[i]);
    }
    return 0;
}

求逆元

模为素数

若(p)为素数，(a)为正整数，且(a)、(p)互质。则有(a^{p-1}equiv 1(mod p))。

quickPow(a, p-2);

模不为素数

void exGCD(LL a, LL b, LL &x, LL &y)
{
    if (!b) x = 1, y = 0;
    else
    {
        exGCD(b, a%b, y, x);
        y -= a/b*x;
    }
}

求一段逆元

#include <bits/stdc++.h>
typedef long long LL;
const int maxn = 20001000;

using namespace std;

LL inv[maxn+10];

int main()
{
    LL n, p;
    scanf("%lld%lld", &n, &p);
    inv[1] = 1;
    for (int i=2; i<=n; ++i)
        inv[i] = (p - p / i) * inv[p % i] % p;
    for (int i=1; i<=n; ++i)
        printf("%lld
", inv[i]);
    return 0;
}

LCA

tarjan离线算法

#include <bits/stdc++.h>
const int maxn = 500000;
const int maxm = 500000;

using namespace std;

int to1[maxn*2+10];
int nxt1[maxn*2+10];
int head1[maxn+10], cnt1 = 0;

void addEdge(int x, int y)
{
    to1[cnt1] = y; nxt1[cnt1] = head1[x];
    head1[x] = cnt1++;
    to1[cnt1] = x; nxt1[cnt1] = head1[y];
    head1[y] = cnt1++;
}

int to2[maxm*2+10];
int w2[maxm*2+10];
int nxt2[maxm*2+10];
int head2[maxm+10], cnt2 = 0;

void addQuery(int a, int b)
{
    to2[cnt2] = b; nxt2[cnt2] = head2[a];
    head2[a] = cnt2++;
    to2[cnt2] = a; nxt2[cnt2] = head2[b];
    head2[b] = cnt2++;
}

int fa[maxn+10];
int vis[maxn+10];

int getFa(int x)
{
    if (fa[x] == x)
        return x;
    return fa[x] = getFa(fa[x]);
}

void LCA(int x, int pre)
{
    for (int i=head1[x]; i!=-1; i=nxt1[i])
    {
        int l = to1[i];
        if (l != pre)
            LCA(l, x);
    }
    //visit the root last
    for (int i=head2[x]; i!=-1; i=nxt2[i])
    {
        int l = to2[i];
        if (vis[l])
            w2[i] = getFa(l);
    }
    fa[x] = pre;
    vis[x] = 1;
}

int main()
{
    int N, M, S;
    scanf("%d%d%d", &N, &M, &S);
    memset(head1, -1, sizeof(head1));
    for (int i=1,x,y; i<=N-1; i++)
    {
        scanf("%d%d", &x, &y);
        addEdge(x, y);
    }
    memset(w2, -1, sizeof(w2));
    memset(head2, -1, sizeof(head2));
    for (int i=1,a,b; i<=M; i++)
    {
        scanf("%d%d", &a, &b);
        addQuery(a, b);
    }
    for (int i=1; i<=N; i++)
        fa[i] = i;
    memset(vis, 0, sizeof(vis));
    LCA(S, S);
    for (int i=0; i<=cnt2-1; i++)
    {
        if (w2[i] != -1)
            printf("%d
", w2[i]);
    }
    return 0;
}

FFT

#include <bits/stdc++.h>
const int maxn = 10000000;
const double PI = acos(-1.0);

using namespace std;

struct tComplex
{
    double x, y;
    tComplex(double xx = 0, double yy = 0)
    {
        x = xx; y = yy;
    }
};
tComplex a[maxn+10];
tComplex b[maxn+10];

tComplex operator + (tComplex a, tComplex b)
{
    return tComplex(a.x + b.x, a.y + b.y);
}

tComplex operator - (tComplex a, tComplex b)
{
    return tComplex(a.x - b.x, a.y - b.y);
}

tComplex operator * (tComplex a, tComplex b)
{
    return tComplex(a.x*b.x - a.y*b.y, a.x*b.y + a.y*b.x);
}

int l, r[maxn+10];
int limit = 1;

void FFT(tComplex A[], int type)
{
    for (int i = 0; i < limit; i++)
    {
        if (i < r[i])
            swap(A[i], A[r[i]]);
    }
    for (int mid = 1; mid < limit; mid <<= 1)
    {
        tComplex Wn(cos(PI/mid), type * sin(PI/mid));
        for (int R = mid << 1, j = 0; j < limit; j += R)
        {
            tComplex w(1, 0);
            for (int k = 0; k < mid; k++, w = w * Wn)
            {
                tComplex x = A[j+k], y = w * A[j+mid+k];
                A[j+k] = x + y;
                A[j+mid+k] = x - y;
            }
        }
    }
}

int main()
{
    int N, M;
    scanf("%d%d", &N, &M);
    for (int i = 0; i <= N; i++)
        scanf("%lf", &a[i].x);
    for (int i = 0; i <= M; i++)
        scanf("%lf", &b[i].x);
    while (limit <= N+M)
        limit <<= 1, l++;
    // get reverse order 0..limit-1
    for (int i = 0; i < limit; i++)
        r[i] = (r[i>>1]>>1) | ((i&1)<<(l-1));
    FFT(a, 1);
    FFT(b, 1);
    for (int i = 0; i < limit; i++)
        a[i] = a[i] * b[i];
    FFT(a, -1);
    for (int i = 0; i <= N+M; i++)
        printf("%d ", (int)(a[i].x / limit + 0.5));
    return 0;
}

Java大数

import java.math.BigInteger;

Scanner in = new Scanner(System.in);
BigInteger x1 = new BigInteger("-11"); //新建一个对象
BigInteger x2 = in.nextBiginteger();//键盘输入
BigInteger y = x1.abs(); //取绝对值
BigInteger y2 = x1.add(y); //x+y
int x3 = y.compareTo(x1); //x和y比较 < == >分别返回 -1,0,1
Boolean x4 = (y.compareTo(x1) <= 0); //判断y是否<=x 0是固定不变的，<=符号可以随意变
BigInteger x5 = y.divide(x1);  // y/x1
BigInteger x6 = y.multiply(x1);// y*x1
double x6 = x2.doubleValue(); //将x2转化为double类型，转化为int用intValue（）
BigInteger x7 = x2.gcd(x1);// 返回abs(x2)和abs(x1)的最大公约数
BigInteger x8 = x2.negate();// 返回-x2
String x9 = x2.toString();// 将x2转化为字符串
BigInteger y3 = x2.reamainder(x1);// 返回x2%x1
BigInteger y4= x1.pow(2);// pow(int exponent) 返回其值为 (this^exponent) 的 BigInteger。

去重离散化

浮点数

double a[maxn+5];//原数组1..n
int b[maxn+5];//离散后数组1..n

struct tdisc
{
    double x;
    int y;
    bool operator<(const tdisc& rhs) const
    {
        return x<rhs.x;
    }
};
tdisc disc[maxn+5];//排序用结构体1..n

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)//编号1..n
    {
        scanf("%lf",a+i);
        disc[i]=(tdisc){a[i],i};
    }
    sort(disc+1,disc+1+n);
    b[disc[1].y]=1;
    for(int i=2,temp=1;i<=n;i++)
    {
        if(fabs(disc[i].x-disc[i-1].x)>eps)
            temp++;
        b[disc[i].y]=temp;
    }
    return 0;
}

整数

int a[maxn+5];//原数组1..n
int b[maxn+5];//离散后数组1..n

struct tdisc
{
    int x;
    int y;
    bool operator<(const tdisc& rhs) const
    {
        return x < rhs.x;
    }
};
tdisc disc[maxn+5];//排序用结构体1..n

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)//编号1..n
    {
        scanf("%d", a+i);
        disc[i] = (tdisc){a[i], i};
    }
    sort(disc+1, disc+1+n);
    b[disc[1].y] = 1;
    for(int i=2, temp=1; i<=n; i++)
    {
        if(disc[i].x > disc[i-1].x)
            temp++;
        b[disc[i].y] = temp;
    }
    for(int i=1; i<=n; i++)
        printf("%d ", b[i]);
    return 0;
}

Cantor展开

//s[0..n-1]中存有n个不同的数，返回值从1开始的序号
int fac[] = {1,1,2,6,24,120,720,5040,40320,362880};
int cantor(int s[],int n)
{
    int sum = 0;
    for(int i = 0; i<n; i++)
    {
        int num = 0;
        for(int j=i+1; j<n; j++)
          if(s[j] < s[i])
            num++;
        sum += (num * fac[n-i-1]);
    }
    return sum + 1;
}

spfa, (O(kE)/O(VE))

//链式前向星
int to[maxm*2+10];//maxm指边最大的数目
int w[maxm*2+10];
int nex[maxm*2+10];
int head[maxn+10], cnt;

int dis[maxn+10];//maxn为点最大的数目
int vis[maxn+10];

queue<int> q; q.push(s);//s为源点
for(int i=1; i<=n; i++)
	dis[i] = inf;
dis[s] = 0; vis[s] = 1;
while(!q.empty())
{
	int p = q.front(); q.pop();
	vis[p] = 0;
	for(int i = head[p]; i != -1; i = nex[i])
	{
        int l = to[i];
        if(dis[l] > dis[p] + w[i])
        {
            dis[l] = dis[p] + w[i];
            if(!vis[l])
            {
                vis[l] = 1;
                q.push(l);
            }
        }
    }
}

dijkstra

一般实现，(O(V^2))

int dis[maxn+10];
int done[maxn+10];

memset(done, 0, sizeof(done));
for(int i=1; i<=n; i++)
	dis[i] = inf;
dis[s] = 0;//s为源点
for(int i=1; i<=n; i++)
{
	int x, m = inf;
	for(int y=1; y<=n; y++)
	{
		if(!done[y] && dis[y]<=m) 
			m = d[x=y];
	}
	done[x] = 1;
	for(int i = head[x]; i != -1; i = nex[i])
	{
		int l = to[i];
		dis[l] = min(dis[l], dis[x]+w[i]);
	}
}

堆优化实现，(O(ElogV))

struct tnode
{
    int d,u;//估计的距离，点的编号
    bool operator < (const tnode& rhs) const
    {
        return d > rhs.d;
    }
};
int dis[maxn+10];//距离
int done[maxn+10];//距离是否已确定

int s = 1;//s为源点
for(int i=1; i<=n; i++)
    dis[i] = inf;
dis[s] = 0;
memset(done, 0, sizeof(done));
priority_queue<tnode> q;
q.push((tnode){0, s});
while(!q.empty())
{
	tnode x = q.top(); q.pop();
	int u = x.u;
	if(done[u])
		continue;
	done[u] = 1;
	for(int i=head[u]; i!=-1; i=nex[i])
	{
		int l = to[i];
		if(dis[l] > dis[u]+w[i])
        {
            dis[l] = dis[u]+w[i];
            q.push((tnode){dis[l], l});
        }
    }
}

带权并查集

struct tnode
{
    int par;
    int rel;//关系
};
tnode node[maxn+10];

int parent(int x)
{
    if(node[x].par != x)
    {
        int temp = node[x].par;
        node[x].par = parent(node[x].par);
        node[x].rel = (node[x].rel+node[temp].rel) % 3;
    }
    return node[x].par;
}

bool unite(int d,int x,int y)
{
    int px=parent(x), py=parent(y);		
	//rel和：px->py应该相当于px->x->y->pyrel
    if(px!=py)
    {
        int temp=-node[x].rel+node[y].rel;
        if(d == 2)
            temp += 1;
        temp = (temp%3+3)%3;//(-1)%3!=2而=-1
        node[px].par = py;
        node[px].rel = temp;
        return true;
    }
    else
    {
        int temp = -node[x].rel+node[y].rel;
        if(d == 2)
            temp += 1;
        temp = (temp%3+3)%3;
        if(temp != 0)
            return false;
        return true;
    }
}

尾声

也许还能用到吧，以后的学习。
算法竞赛，就此终了。
算法的学习，更广阔的学习，我仍在路上！