BZOJ2001 [Hnoi2010]City 城市建设 【CDQ分治 + kruskal】

题目链接

BZOJ2001

题解

CDQ分治神题。。。

难想难写。。

比较朴素的思想是对于每个询问都求一遍\(BST\)，这样做显然会爆

考虑一下时间都浪费在了什么地方

我们每次求\(BST\)实际上就只有一条边不同，我们实际浪费了很多时间在处理相同的边上

那就考虑分治

对于一个待修改的边集，我们将其权值全部设为\(-\infty\)，跑一遍\(BST\)，此时其它边如果被选中，说明这些边在单独询问时也一定会被选，将这些边连的点缩点

同样，对于一个待修改的边集，我们将其权值全部设为\(\infty\)，跑一遍\(BST\)，此时其它边没被选中，说明这些边在单独询问时也一定不会被选，将这些边删掉

这样就可以\(A\)了

复杂度我也不知道是什么

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 50005,maxm = 50005,INF = 0x3fffffff;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
struct EDGE{int i,a,b,w;}e[50][maxm],d[maxm],b[maxm];
struct Que{int u,v;}q[maxm];
inline bool operator <(const EDGE& a,const EDGE& b){
    return a.w < b.w;
}
LL ans[maxm];
int n,m,Q,sum[50],w[maxm],id[maxm],a[maxm];
int pre[maxm];
int find(int u){return u == pre[u] ? u : pre[u] = find(pre[u]);}
void clear(int t){
    for (int i = 1; i <= t; i++){
        pre[d[i].a] = d[i].a;
        pre[d[i].b] = d[i].b;
    }
}
void comb(int& t,LL& Ans){
    clear(t); int tmp = 0,fa,fb;
    sort(d + 1,d + 1 + t);
    REP(i,t){
        fa = find(d[i].a); fb = find(d[i].b);
        if (fa != fb){
            pre[fb] = fa;
            b[++tmp] = d[i];
        }
    }
    REP(i,tmp) {
        pre[b[i].a] = b[i].a;
        pre[b[i].b] = b[i].b;
    }
    REP(i,tmp) if (b[i].w != -INF){
        fa = find(b[i].a); fb = find(b[i].b);
        if (fa != fb){
            pre[fb] = fa;
            Ans += b[i].w;
        }
    }
    tmp = 0;
    REP(i,t){
        fa = find(d[i].a); fb = find(d[i].b);
        if (fa != fb){
            b[++tmp] = d[i];
            id[d[i].i] = tmp;
            b[tmp].a = find(b[tmp].a);
            b[tmp].b = find(b[tmp].b);
        }
    }
    REP(i,tmp) d[i] = b[i];
    t = tmp;
}
void rd(int& t){
    clear(t); int tmp = 0,fa,fb;
    sort(d + 1,d + 1 + t);
    REP(i,t){
        fa = find(d[i].a); fb = find(d[i].b);
        if (fa != fb){
            pre[fb] = fa;
            b[++tmp] = d[i];
        }
        else if (d[i].w == INF){
            b[++tmp] = d[i];
        }
    }
    for (int i = 1; i <= tmp; i++) d[i] = b[i];
    t = tmp;
}
void solve(int l,int r,int now,LL Ans){
    int t = sum[now];
    if (l == r) a[q[l].u] = q[l].v;   //原标号边权值
    for (int i = 1; i <= t; i++)
        e[now][i].w = a[e[now][i].i];  //边赋值
    for (int i = 1; i <= t; i++)
        d[i] = e[now][i],id[d[i].i] = i;  //新边对应旧边位置
    if (l == r){
        ans[l] = Ans; clear(t);
        sort(d + 1,d + 1 + t);
        int fa,fb;
        for (int i = 1; i <= t; i++){
            fa = find(d[i].a); fb = find(d[i].b);
            if (fa != fb){
                pre[fb] = fa; ans[l] += d[i].w;
            }
        }
        return;
    }
    for (int i = l; i <= r; i++) d[id[q[i].u]].w = -INF;
    comb(t,Ans);
    for (int i = l; i <= r; i++) d[id[q[i].u]].w = INF;
    rd(t);
    REP(i,t) e[now + 1][i] = d[i];
    sum[now + 1] = t;
    int mid = l + r >> 1;
    solve(l,mid,now + 1,Ans);
    solve(mid + 1,r,now + 1,Ans);
}
int main(){
    n = read(); m = read(); Q = read();
    for (int i = 1; i <= m; i++){
        e[0][i].i = i;
        e[0][i].a = read();
        e[0][i].b = read();
        a[i] = e[0][i].w = read();
    }
    for (int i = 1; i <= Q; i++){
        q[i].u = read();
        q[i].v = read();
    }
    sum[0] = m;
    solve(1,Q,0,0);
    for (int i = 1; i <= Q; i++)
        printf("%lld\n",ans[i]);
    return 0;
}

原文地址：https://www.cnblogs.com/Mychael/p/9065839.html