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+;
 const int N = 1e5 + ;
 struct Node{
     int u, v, c, id;
     bool operator < (const Node & x) const{
         return c < x.c;
     }
 }e[][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]] = ;
     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 = ; 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+, f++tot);
     int u, v, zz = ;
     for(int i = ; 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 = ; i <= zz; i++){
         u = Find(g[i].u); v = Find(g[i].v);
         pre[u] = v;
     }
     zz = ;
     for(int i = ; 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+, f++tot);
     int u, v, zz = ;
     for(int i = ; 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 = ; 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+, f++tot);
         int u, v;
         for(int i = ; 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 = ; i <= tot; i++) e[now+][i] = f[i];
     int mid = l+r >> ;
     cdq(l, mid, now+, tot, sum);
     cdq(mid+, r, now+, tot, sum); }
 int main(){
     int n, m, q;
     scanf("%d%d%d", &n, &m, &q);
     for(int i = ; i <= m; i++){
         scanf("%d%d%d", &e[][i].u, &e[][i].v, &e[][i].c);
         e[][i].id = i;
         ct[i] = e[][i].c;
     }
     for(int i = ; i <= q; i++)
         scanf("%d%d", &a[i], &b[i]);
     cdq(,q,,m,);
     for(int i = ; i <= q; ++i)
         printf("%lld\n", ans[i]);
     return ;
 }