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：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define lch(x) tr[x].son[0] 12 #define rch(x) tr[x].son[1] 13 #define max3(a,b,c) max(a,max(b,c)) 14 #define min3(a,b,c) min(a,min(b,c)) 15 typedef pair<int,int> pll; 16 const int inf = 0x3f3f3f3f; 17 const LL INF = 0x3f3f3f3f3f3f3f3f; 18 const LL mod = (int)1e9+7; 19 const int N = 1e5 + 100; 20 struct Node{ 21 int u, v, c, id; 22 bool operator < (const Node & x) const{ 23 return c < x.c; 24 } 25 }e[20][N], f[N], g[N]; 26 int a[N], b[N], ct[N], mapid[N]; 27 int pre[N]; 28 int to[N]; 29 void link(int u, int v){ 30 mapid[to[v]] = 0; 31 mapid[u] = v; 32 to[v] = u; 33 } 34 int Find(int x){ 35 if(x == pre[x]) return x; 36 return pre[x] = Find(pre[x]); 37 } 38 LL ans[N]; 39 void Clear(int tot){ 40 for(int i = 1; i <= tot; i++){ 41 pre[f[i].u] = f[i].u; 42 pre[f[i].v] = f[i].v; 43 } 44 } 45 void contraction(int &tot, LL &sum){ 46 Clear(tot); 47 sort(f+1, f+1+tot); 48 int u, v, zz = 0; 49 for(int i = 1; i <= tot; i++){ 50 u = Find(f[i].u), v = Find(f[i].v); 51 if(u != v){ 52 pre[u] = v; 53 if(f[i].c != -inf){ 54 sum += f[i].c; 55 g[++zz] = f[i]; 56 } 57 } 58 } 59 Clear(tot); 60 for(int i = 1; i <= zz; i++){ 61 u = Find(g[i].u); v = Find(g[i].v); 62 pre[u] = v; 63 } 64 zz = 0; 65 for(int i = 1; i <= tot; i++){ 66 u = Find(f[i].u), v = Find(f[i].v); 67 if(u != v){ 68 f[++zz] = f[i]; 69 f[zz].u = u; 70 f[zz].v = v; 71 mapid[f[i].id] = zz; 72 } 73 } 74 tot = zz; 75 return ; 76 } 77 void reduction(int &tot){ 78 Clear(tot); 79 sort(f+1, f+1+tot); 80 int u, v, zz = 0; 81 for(int i = 1; i <= tot; i++){ 82 u = Find(f[i].u); v = Find(f[i].v); 83 if(u != v){ 84 pre[u] = v; 85 f[++zz] = f[i]; 86 } 87 else if(f[i].c == inf) 88 f[++zz] = f[i]; 89 } 90 tot = zz; 91 return ; 92 } 93 void cdq(int l, int r, int now, int tot, LL sum){ 94 if(l == r) ct[a[l]] = b[l]; 95 for(int i = 1; i <= tot; i++){ 96 e[now][i].c = ct[e[now][i].id]; 97 mapid[e[now][i].id] = i; 98 //link(e[now][i]) 99 f[i] = e[now][i]; 100 } 101 if(l == r){ 102 ans[l] = sum; 103 Clear(tot); 104 sort(f+1, f+1+tot); 105 int u, v; 106 for(int i = 1; i <= tot; i++){ 107 u = Find(f[i].u), v = Find(f[i].v); 108 if(u != v){ 109 pre[u] = v; 110 ans[l] += f[i].c; 111 } 112 } 113 return ; 114 } 115 for(int i = l; i <= r; i++) f[mapid[a[i]]].c = -inf; 116 contraction(tot, sum); 117 for(int i = l; i <= r; i++) f[mapid[a[i]]].c = inf; 118 reduction(tot); 119 for(int i = 1; i <= tot; i++) e[now+1][i] = f[i]; 120 int mid = l+r >> 1; 121 cdq(l, mid, now+1, tot, sum); 122 cdq(mid+1, r, now+1, tot, sum); 123 124 } 125 int main(){ 126 int n, m, q; 127 scanf("%d%d%d", &n, &m, &q); 128 for(int i = 1; i <= m; i++){ 129 scanf("%d%d%d", &e[0][i].u, &e[0][i].v, &e[0][i].c); 130 e[0][i].id = i; 131 ct[i] = e[0][i].c; 132 } 133 for(int i = 1; i <= q; i++) 134 scanf("%d%d", &a[i], &b[i]); 135 cdq(1,q,0,m,0); 136 for(int i = 1; i <= q; ++i) 137 printf("%lld ", ans[i]); 138 return 0; 139 }