uestc summer training #3 线段树优化建边

Posted 2020-12-05 aragaki

A

如果把边数缩小到n^2可以接受的话 就是一个最小点基的裸题

但是这里可能有n^2条边所以我们需要线段树优化建边 然后再求出SCC

扣掉不包含原始n个节点的SCC或者把除叶子节点外线段树上的点权设为inf 然后跑最小点基

#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
typedef pair<int, int> P;
const int N = 1000010, M = 8000000;
int n, m, i, j, x, y;
long long ans;
struct E
{
        int p, r, c;
} a[N];
P b[N];
int root, l[N], r[N], tot;
set<P>T[N];
int g[2][N], nxt[2][M], v[2][M], ed, f[N], q[N], t, vis[N], ban[N];
inline void add(int x, int y)
{
        v[0][++ed] = y;
        nxt[0][ed] = g[0][x];
        g[0][x] = ed;
        v[1][ed] = x;
        nxt[1][ed] = g[1][y];
        g[1][y] = ed;
}
inline void ADD(int x, int y)
{
        v[1][++ed] = y;
        nxt[1][ed] = g[1][x];
        g[1][x] = ed;
}
int build(int a, int b)
{
        int x;
        if (a == b)  //如果该点是叶子节点的话 值就为下标
        {
                x =::b[a].second;
        }
        else //否则的话 就给该节点一个标号
        {
                x = ++tot;
        }
        if (a == b)
        {
                return x;
        }
        int mid = (a + b) >> 1;
        l[x] = build(a, mid);
        r[x] = build(mid + 1, b);
        add(x, l[x]);
        add(x, r[x]);
        return x;
}
void ins(int x, int a, int b, int c, int d, int p)
{
        if (c <= a && b <= d)
        {
                add(p, x);  //p是不会变的 如果满足条件的话就把p和x节点连上一条边
                return;
        }
        int mid = (a + b) >> 1;
        if (c <= mid)
        {
                ins(l[x], a, mid, c, d, p);
        }
        if (d > mid)
        {
                ins(r[x], mid + 1, b, c, d, p);
        }
}
inline int askl(int x) //min >=x
{
        int l = 1, r = n, mid, t;
        while (l <= r)
        {
                mid = (l + r) >> 1;
                if (b[mid].first >= x)
                {
                        r = (t = mid) - 1;
                }
                else
                {
                        l = mid + 1;
                }
        }
        return t;
}
inline int askr(int x) //max <=x
{
        int l = 1, r = n, mid, t;
        while (l <= r)
        {
                mid = (l + r) >> 1;
                if (b[mid].first <= x)
                {
                        l = (t = mid) + 1;
                }
                else
                {
                        r = mid - 1;
                }
        }
        return t;
}
void dfs1(int x)
{
        vis[x] = 1;
        for (int i = g[0][x]; i; i = nxt[0][i])
                if (!vis[v[0][i]])
                {
                        dfs1(v[0][i]);
                }
        q[++t] = x;
}
void dfs2(int x, int y)
{
        vis[x] = 0;
        f[x] = y;
        for (int i = g[1][x]; i; i = nxt[1][i])
                if (vis[v[1][i]])
                {
                        dfs2(v[1][i], y);
                }
}
void dfs3(int x)
{
        if (ban[x])
        {
                return;
        }
        ban[x] = 1;
        for (int i = g[1][x]; i; i = nxt[1][i])
        {
                dfs3(v[1][i]);
        }
}
inline void solve(int x)
{
        if (vis[x])
        {
                return;
        }
        vis[x] = 1;
        for (int i = g[1][x]; i; i = nxt[1][i])
        {
                dfs3(v[1][i]);
        }
}
int main()
{
        scanf("%d%d", &n, &m);
        for (i = 1; i <= n; i++)
        {
                scanf("%d%d%d", &a[i].p, &a[i].r, &a[i].c);
                b[i] = P(a[i].p, i);
        }
        sort(b + 1, b + n + 1); //根据每个点的位置进行排序
        tot = n; //初始会有n个节点
        root = build(1, n);  //建立线段树并对线段树上的节点进行赋值
        for (i = 1; i <= n; i++)
        {
                int l = askl(a[i].p - a[i].r); //二分得到最左边炸到的节点
                int r = askr(a[i].p + a[i].r); //二分得到最右边炸到的节点
                ins(root, 1, n, l, r, i); //把该节点和线段树上范围为子区间的节点连一条边
        }
        for (t = 0, i = 1; i <= tot; i++)
                if (!vis[i])
                {
                        dfs1(i);
                }
        for (i = tot; i; i--)
                if (vis[q[i]])
                {
                        dfs2(q[i], q[i]);
                }
        ed = 0;  //ed为SCC的边总数
        for (i = 1; i <= tot; i++)  //SCC前向星初始化head数组
        {
                g[1][i] = 0;
        }
        for (i = 1; i <= tot; i++)
                for (j = g[0][i]; j; j = nxt[0][j])
                        if (f[i] != f[v[0][j]]) //不同SCC之间建边
                        {
                                ADD(f[i], f[v[0][j]]);
                        }
        for (i = 1; i <= n; i++)
        {
                solve(f[i]);
        }
        for (i = 1; i <= n; i++)
                if (!ban[f[i]]) //如果f[i]这个SCC是合法的话 就插入该点的一个值
                {
                        T[f[i]].insert(P(a[i].c, i));
                }
        for (i = 1; i <= tot; i++)
                if (!ban[i] && f[i] == i) //如果这个SCC合法且这个SCC的入度是0的话 就把这个SCC内最小的点权值加上
                {
                        ans += T[i].begin()->first;
                }
        while (m--)
        {
                scanf("%d%d", &x, &y);
                if (!ban[f[x]])  //
                {
                        ans -= T[f[x]].begin()->first;
                        T[f[x]].erase(P(a[x].c, x));
                        T[f[x]].insert(P(a[x].c = y, x));
                        ans += T[f[x]].begin()->first;
                }
                printf("%lld
", ans);
        }
}

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