bzoj 2001 CITY 城市建设 cdq分治

题目传送门

题解：

对整个修改的区间进行分治。对于当前修改区间来说，我们对整幅图中将要修改的边权都先改成-inf，跑一遍最小生成树，然后对于一条树边并且他的权值不为-inf，那么这条边一定就是树边了。然后我们把这些点都缩成一个点。然后，我们继续对当前修改区间来说，我们把要修改的边的边权都修改成inf，跑一遍最小生成树，然后对于一条非树边来说，他的边权不为inf，那么这条边一点是非树边了，然后我们每层缩点，减边，这样图就会越来越小，然后当l == r的时候，我们还原修改操作，最后把跑最小生成树计算答案。

一道神奇的cdq题目。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
struct Node{
    int u, v, c, id;
    bool operator < (const Node & x) const{
        return c < x.c;
    }
}e[20][N], f[N], g[N];
int a[N], b[N], ct[N], mapid[N];
int pre[N];
int to[N];
void link(int u, int v){
    mapid[to[v]] = 0;
    mapid[u] = v;
    to[v] = u;
}
int Find(int x){
    if(x == pre[x]) return x;
    return pre[x] = Find(pre[x]);
}
LL ans[N];
void Clear(int tot){
    for(int i = 1; i <= tot; i++){
        pre[f[i].u] = f[i].u;
        pre[f[i].v] = f[i].v;
    }
}
void contraction(int &tot, LL &sum){
    Clear(tot);
    sort(f+1, f+1+tot);
    int u, v, zz = 0;
    for(int i = 1; i <= tot; i++){
        u = Find(f[i].u), v = Find(f[i].v);
        if(u != v){
            pre[u] = v;
            if(f[i].c != -inf){
                sum += f[i].c;
                g[++zz] = f[i];
            }
        }
    }
    Clear(tot);
    for(int i = 1; i <= zz; i++){
        u = Find(g[i].u); v = Find(g[i].v);
        pre[u] = v;
    }
    zz = 0;
    for(int i = 1; i <= tot; i++){
        u = Find(f[i].u), v = Find(f[i].v);
        if(u != v){
            f[++zz] = f[i];
            f[zz].u = u;
            f[zz].v = v;
            mapid[f[i].id] = zz;
        }
    }
    tot = zz;
    return ;
}
void reduction(int &tot){
    Clear(tot);
    sort(f+1, f+1+tot);
    int u, v, zz = 0;
    for(int i = 1; i <= tot; i++){
        u = Find(f[i].u); v = Find(f[i].v);
        if(u != v){
            pre[u] = v;
            f[++zz] = f[i];
        }
        else if(f[i].c == inf)
            f[++zz] = f[i];
    }
    tot = zz;
    return ;
}
void cdq(int l, int r, int now, int tot, LL sum){
    if(l == r) ct[a[l]] = b[l];
    for(int i = 1; i <= tot; i++){
        e[now][i].c = ct[e[now][i].id];
        mapid[e[now][i].id] = i;
        //link(e[now][i])
        f[i] = e[now][i];
    }
    if(l == r){
        ans[l] = sum;
        Clear(tot);
        sort(f+1, f+1+tot);
        int u, v;
        for(int i = 1; i <= tot; i++){
            u = Find(f[i].u), v = Find(f[i].v);
            if(u != v){
                pre[u] = v;
                ans[l] += f[i].c;
            }
        }
        return ;
    }
    for(int i = l; i <= r; i++) f[mapid[a[i]]].c = -inf;
    contraction(tot, sum);
    for(int i = l; i <= r; i++) f[mapid[a[i]]].c = inf;
    reduction(tot);
    for(int i = 1; i <= tot; i++) e[now+1][i] = f[i];
    int mid = l+r >> 1;
    cdq(l, mid, now+1, tot, sum);
    cdq(mid+1, r, now+1, tot, sum);

}
int main(){
    int n, m, q;
    scanf("%d%d%d", &n, &m, &q);
    for(int i = 1; i <= m; i++){
        scanf("%d%d%d", &e[0][i].u, &e[0][i].v, &e[0][i].c);
        e[0][i].id = i;
        ct[i] = e[0][i].c;
    }
    for(int i = 1; i <= q; i++)
        scanf("%d%d", &a[i], &b[i]);
    cdq(1,q,0,m,0);
    for(int i = 1; i <= q; ++i)
        printf("%lld\n", ans[i]);
    return 0;
}