[树链剖分+MST] CF609E. Minimum spanning tree for each edge

题目链接：Minimum spanning tree for each edge

题意：给一个无向图，n个点，m条边，对任意边edge[i]，求出包含有边edge[i]的最小生成树。

做法：考虑MST的性质，对任意两点u,v一定有且只有一条路径，当边[u,v]不是MST边的时候，加入边[u,v]便会形成环，于是在u到v的路径上找一条权值最大的边删去再加入边[u,v]，之前环上的任意两点仍然连通。

u到v路径的最大值用树链剖分维护，LCA大概也可以，不过做的时候顺便套了树剖的板子。

最开始求一次原始的MST，建立树剖后，就可以很方便地求出每个边的答案啦。

#include
#define ll long long int
using namespace std;
const int N = 200005;
struct eg{
    int u, v, nex, val, id;
    eg(){}
    eg(int _u, int _v, int _val, int _nex) { u = _u, v = _v, val = _val, nex = _nex; }
    bool operator < (eg a) const{
        return val < a.val;
    }
}edg[N<<2], MST[N];
ll ans[N] = {0};
int sav[N][3], scnt;
int fir[N], ecnt, root;
int n, m;
void add(int u, int v, int val){
    edg[ecnt] = eg(u, v, val, fir[u]);
    fir[u] = ecnt++;
    edg[ecnt] = eg(v, u, val, fir[v]);
    fir[v] = ecnt++;
}

////////////////////////////////////////////////////树链剖分
int fa[N], dep[N], siz[N], son[N];
int tree[N<<2];
void dfs(int rt){
    siz[rt] = 1, son[rt] = 0;
    for(int k = fir[rt]; k != -1; k = edg[k].nex){
        if(edg[k].v != fa[rt]){
            fa[edg[k].v] = rt;
            dep[edg[k].v] = dep[rt] + 1;
            dfs(edg[k].v);
            if(siz[edg[k].v] > siz[son[rt]]) son[rt] = edg[k].v;
            siz[rt] += siz[edg[k].v];
        }
    }
}
int w[N], top[N],  cnt; //w[]为到线段树的映射
void dfs2(int rt, int tp){
    w[rt] = ++cnt; top[rt] = tp;
    if(son[rt]) dfs2(son[rt], top[rt]);
    for(int k = fir[rt]; k != -1; k = edg[k].nex){
        if(edg[k].v != son[rt] && edg[k].v != fa[rt]){ 
            dfs2(edg[k].v, edg[k].v);
        }
    }
}
void update(int rt, int l, int r, int pos, int val){
    if(l == r){
        tree[rt] = val;
        return;
    }
    int mid = (l+r) >> 1;
    if(pos <= mid) update(rt<<1, l, mid, pos, val);
    else update(rt<<1|1, mid+1, r, pos, val);
    tree[rt] = max(tree[rt<<1], tree[rt<<1|1]);
}
void init(){
    memset(siz, 0, sizeof(siz));
    memset(tree, 0, sizeof(tree));
    root = (n+1)/2;
    fa[root] = cnt = dep[root] = 0;
    dfs(root);
    dfs2(root, root);
    for(int i = 1; i <= scnt; ++i){
        if(dep[sav[i][0]] > dep[sav[i][1]]) swap(sav[i][0], sav[i][1]);
        update(1, 1, cnt, w[sav[i][1]], sav[i][2]);
    }
}
int maxi(int rt, int l, int r, int ql, int qr){
    if( ql > r || qr < l) return 0;
    if(ql <= l && qr >= r) return tree[rt];
    int mid = (l+r) >> 1;
    return max(maxi(rt<<1, l, mid, ql, qr), maxi(rt<<1|1, mid+1, r, ql, qr));
}
int Treemax(int va, int vb){
    int f1 = top[va], f2 = top[vb], tmp = 0;
    while(f1 != f2){
        if(dep[f1] < dep[f2]){
            swap(f1, f2);
            swap(va, vb);
        }
        tmp = max(tmp, maxi(1, 1, cnt, w[f1], w[va]));
        va = fa[f1], f1 = top[va];
    }
    if(va == vb) return tmp;
    if(dep[va] > dep[vb]) swap(va, vb);
    return max(tmp, maxi(1, 1, cnt, w[son[va]], w[vb]));
}
///////////////////////////////////////////////////////

//////////////////////////////////////////////////////并查集
int seed[N];
int find(int x){
    return seed[x] < 0? x : seed[x] = find(seed[x]);
}
int join(int a, int b){
    a = find(a), b = find(b);
    if(a == b) return 0;
    if(seed[a] > seed[b]) seed[b] += seed[a], seed[a] = b;
    else seed[a] += seed[b], seed[b] = a;
    return 1;
}
/////////////////////////////////////////////////////

int main(){
    memset(seed, -1, sizeof(seed));
    memset(fir, -1, sizeof(fir));
    ecnt = scnt = 0;

    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; ++i){
        scanf("%d %d %d", &MST[i].u, &MST[i].v, &MST[i].val);
        MST[i].id = i+1;
    }
    /////////////Kruskal求MST
    sort(MST, MST+m);
    for(int i = 0; i < m; ++i){
        if(join(MST[i].u, MST[i].v)) {
            ans[0] += MST[i].val;
            //用MST里的边做树链剖分
            add(MST[i].u, MST[i].v, MST[i].val);
            sav[++scnt][0] = MST[i].u, sav[scnt][1] = MST[i].v, sav[scnt][2] = MST[i].val;
        }
    }
    //////////////

    //树链剖分go
    init();
    ////////////

    //然后就可以求解啦
    for(int i = 0; i < m; ++i) ans[MST[i].id] = ans[0] + MST[i].val - Treemax(MST[i].u, MST[i].v);
    for(int i = 1; i <= m; ++i) printf("%lld\n", ans[i]);
}