LCA+最小生成树 Codeforces609E Minimum spanning tree for each edge

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题意：给一个图，有m条边n个点，如果对于一个最小生成树中要求必须包括第i条边，那么最小生成树的权值总和最小是多少

思路：求出最小生成树，然后对于m条边相当于m次查询，每次查询的时候，相当于求出在最小生成树中(u,v)路径上的边权最大值，那么新添加了一条边，就要把这条最大值的边删掉。所以题目转换成了，求路径上边权最大值。可以用LCA来做，也可以用树链剖分来维护。

LCA维护

#include<map>#include<set>#include<cmath>#include<ctime>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define fuck(x) cout<<"["<<x<<"]"#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 2e5 + 5;const int MS = 4e5 + 5;const int M = 25;//n的logconst int INF = 0x3f3f3f3f;struct Edge {    int u, v, nxt, cost, id;    bool operator<(const Edge &P) const {        return cost < P.cost;    }} E[MS], A[MX];int rear, Head[MX];void edge_init() {    rear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v, int cost) {    E[rear].u = u;    E[rear].v = v;    E[rear].cost = cost;    E[rear].nxt = Head[u];    Head[u] = rear++;}LL mincost, ans[MX];int n, m, P[MX];int find(int x) {    return P[x] == x ? x : (P[x] = find(P[x]));}void MST() {    mincost = 0;    for(int i = 1; i <= n; i++) P[i] = i;    for(int i = 1; i <= m; i++) {        int p1 = find(A[i].u), p2 = find(A[i].v);        if(p1 != p2) {            P[p1] = p2;            edge_add(A[i].u, A[i].v, A[i].cost);            edge_add(A[i].v, A[i].u, A[i].cost);            mincost += A[i].cost;        }    }}int dep[MX], fa[MX][M], MAX[MX][M];void DFS(int u, int _dep, int _fa) {    dep[u] = _dep; fa[u][0] = _fa;    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(v == _fa) {            MAX[u][0] = E[i].cost;            continue;        }        DFS(v, _dep + 1, u);    }}void presolve() {    DFS(1, 0, 1);    for(int i = 1; i < M; i++) {        for(int j = 1; j <= n; j++) {            fa[j][i] = fa[fa[j][i - 1]][i - 1];            MAX[j][i] = max(MAX[j][i - 1], MAX[fa[j][i - 1]][i - 1]);        }    }}int LCA(int u, int v) {    int ret = 0;    while(dep[u] != dep[v]) {        if(dep[u] < dep[v]) swap(u, v);        int d = dep[u] - dep[v];        for(int i = 0; i < M; i++) {            if(d >> i & 1) {                ret = max(ret, MAX[u][i]);                u = fa[u][i];            }        }    }    if(u == v) return ret;    for(int i = M - 1; i >= 0; i--) {        if(fa[u][i] != fa[v][i]) {            ret = max(ret, MAX[u][i]);            ret = max(ret, MAX[v][i]);            u = fa[u][i];            v = fa[v][i];        }    }    return max(ret, max(MAX[u][0], MAX[v][0]));}int main() {    //FIN;    while(~scanf("%d%d", &n, &m)) {        edge_init();        for(int i = 1; i <= m; i++) {            A[i].id = i;            scanf("%d%d%d", &A[i].u, &A[i].v, &A[i].cost);        }        sort(A + 1, A + 1 + m);        MST();        presolve();        for(int i = 1; i <= m; i++) {            ans[A[i].id] = mincost - LCA(A[i].u, A[i].v) + A[i].cost;        }        for(int i = 1; i <= m; i++) {            printf("%I64d\n", ans[i]);        }    }    return 0;}

树链剖分维护

#include<map>#include<set>#include<cmath>#include<ctime>#include<stack>#include<queue>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define fuck(x) cout<<"["<<x<<"]"#define FIN freopen("input.txt","r",stdin)#define FOUT freopen("output.txt","w+",stdout)using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 2e5 + 5;const int MS = 4e5 + 5;const int INF = 0x3f3f3f3f;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct Edge {    int u, v, nxt, cost, id;    bool operator<(const Edge &P) const {        return cost < P.cost;    }} E[MS], A[MS];int _rear, Head[MX];void edge_init() {    _rear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v, int cost) {    E[_rear].u = u;    E[_rear].v = v;    E[_rear].cost = cost;    E[_rear].nxt = Head[u];    Head[u] = _rear++;}bool cmp(Edge a, Edge b) {    return a.id < b.id;}int n, m, P[MX];LL mincost;int find(int x) {    return P[x] == x ? x : (P[x] = find(P[x]));}void MST_solve() {    mincost = 0;    edge_init();    sort(A + 1, A + 1 + m);    for(int i = 1; i <= n; i++) P[i] = i;    for(int i = 1; i <= m; i++) {        int p1 = find(A[i].u), p2 = find(A[i].v);        if(p1 != p2) {            P[p1] = p2;            mincost += A[i].cost;            edge_add(A[i].u, A[i].v, A[i].cost);            edge_add(A[i].v, A[i].u, A[i].cost);            //printf("[%d,%d,%d]",A[i].u,A[i].v,A[i].cost);        }    }}int MAX[MX << 2], TA[MX];void push_up(int rt) {    MAX[rt] = max(MAX[rt << 1], MAX[rt << 1 | 1]);}void build(int l, int r, int rt) {    if(l == r) {        MAX[rt] = TA[l];        return;    }    int m = (l + r) >> 1;    build(lson);    build(rson);    push_up(rt);}int query(int L, int R, int l, int r, int rt) {    if(L <= l && r <= R) {        return MAX[rt];    }    int m = (l + r) >> 1, ret = -INF;    if(L <= m) ret = max(ret, query(L, R, lson));    if(R > m) ret = max(ret, query(L, R, rson));    return ret;}int fa[MX], top[MX], siz[MX], son[MX], dep[MX], id[MX], rear;void DFS1(int u, int f, int d) {    fa[u] = f; dep[u] = d;    son[u] = 0; siz[u] = 1;    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(v == f) continue;        DFS1(v, u, d + 1);        siz[u] += siz[v];        if(siz[son[u]] < siz[v]) {            son[u] = v;        }    }}void DFS2(int u, int tp) {    top[u] = tp;    id[u] = ++rear;    if(son[u]) DFS2(son[u], tp);    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(v == fa[u] || v == son[u]) continue;        DFS2(v, v);    }}void HLD_presolve() {    rear = 0;    DFS1(1, 0, 1);    DFS2(1, 1);    for(int i = 0; i < 2 * (rear - 1); i += 2) {        int u = E[i].u, v = E[i].v;        if(dep[u] < dep[v]) swap(u, v);        TA[id[u]] = E[i].cost;    }    TA[1] = -INF;    build(1, rear, 1);}int HLD_query(int u, int v) {    int tp1 = top[u], tp2 = top[v], ans = -INF;    while(tp1 != tp2) {        if(dep[tp1] < dep[tp2]) {            swap(u, v);            swap(tp1, tp2);        }        ans = max(ans, query(id[tp1], id[u], 1, rear, 1));        u = fa[tp1]; tp1 = top[u];    }    if(u == v) return ans;    if(dep[u] > dep[v]) swap(u, v);    ans = max(ans, query(id[son[u]], id[v], 1, rear, 1));    return ans;}int main() {    //FIN;    while(~scanf("%d%d", &n, &m)) {        for(int i = 1; i <= m; i++) {            A[i].id = i;            scanf("%d%d%d", &A[i].u, &A[i].v, &A[i].cost);        }        MST_solve();        HLD_presolve();        sort(A + 1, A + 1 + m, cmp);        for(int i = 1; i <= m; i++) {            int u = A[i].u, v = A[i].v, cost = A[i].cost;            printf("%I64d\n", mincost + cost - HLD_query(u, v));        }    }    return 0;}