HDU

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6074

题目大意：Q住在节点1，他想要连电话线，而在S(ai,bi)∪S(ci,di)之中的任意两个节点都可以连接一条电话线，花费为w，问连得最多节点的最小花费是多少

解题思路：问题转换一下就是让图连通的最小花费是多少，这就转化成好了最小生成树的问题，我们先根据w值排序，处理出S(ai,bi)最小生成树，再出理出S(ci,di)的最小生成树，再把两个合并，但是在他们各自处理的过程中，为了避免S(ai,bi)和S(ci,di)可能部分重合的情况，可以在处理过程中，把S(ai,bi)的所有节点都指向lca(ai,bi)，那么在处理S(ci,di)时，发现其中某个点指向的点的深度大于lca(ci,di)，那么这个点就是处理过的，就不处理。关于Lca树链剖分法，可以参考http://blog.csdn.net/nightmare_ak/article/details/77164374

AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int MAXN = 100000 + 5;struct Edge{    int v, next;    Edge(int v = 0, int next = 0) :v(v), next(next) {}}edge[MAXN << 1];struct Node{    int a, b, c, d;    LL w;    Node(int a = 0, int b = 0, int c = 0, int d = 0, LL w = 0)        :a(a), b(b), c(c), d(d), w(w) {}    bool operator<(const Node& a)const    {        return w < a.w;    }}node[MAXN];int son[MAXN], bulk[MAXN], dep[MAXN], ft[MAXN], top[MAXN];//求lcaint head[MAXN], edgenum;//链接表int anc[MAXN], tag[MAXN];//最小生成树int cnt[MAXN];//答案储存LL cost[MAXN];void toInit(int n){    edgenum = 0;    for (int i = 1;i <= n;++i)    {        anc[i] = tag[i] = i;        cnt[i] = 1;//memset只能处理-1，0        cost[i] = 0;        head[i] = -1;    }}void toAdd(int u, int v){    edge[edgenum] = Edge(v, head[u]);    head[u] = edgenum++;}void toDfs1(int u, int f, int tier){    son[u] = 0;dep[u] = tier;ft[u] = f;bulk[u] = 1;    for (int i = head[u];i != -1;i = edge[i].next)    {        int v = edge[i].v;        if (v == f) continue;        toDfs1(v, u, tier + 1);        bulk[u] += bulk[v];        if (bulk[v] > bulk[son[u]])            son[u] = v;    }}void toDfs2(int u, int f, int rt){    top[u] = rt;    for (int i = head[u];i != -1;i = edge[i].next)    {        int v = edge[i].v;        if (v == f) continue;        if (v == son[u])            toDfs2(v, u, rt);        else            toDfs2(v, u, v);    }}int toLca(int x, int y){    while (top[x] != top[y])    {        if (dep[top[x]] < dep[top[y]])            swap(x, y);        x = ft[top[x]];    }    return dep[x] < dep[y] ? x : y;}int toFind(int x,int F[]){    return F[x] == x ? x : F[x] = toFind(F[x], F);}void toUnion(int x, int y, LL w){    x = toFind(x, anc);    y = toFind(y, anc);    if (x != y)    {        anc[y] = x;        cost[x] += cost[y] + w;        cnt[x] += cnt[y];    }}void toSolve(int r, int x, LL w){    while (1)    {        x = toFind(x, tag);        if (dep[x] <= dep[r])//判断x是否已经合并过            break;        toUnion(x, ft[x], w);        tag[x] = ft[x];//不断地向祖先r合并    }}void divideCal(int x, int y, LL w){    int r = toLca(x, y);    toSolve(r, y, w);    toSolve(r, x, w);}int main(){    int t;scanf("%d", &t);    while (t--)    {        int n, m;        scanf("%d%d", &n, &m);        toInit(n);        for (int i = 1;i <= n - 1;++i)        {            int u, v;            scanf("%d%d", &u, &v);            toAdd(u, v);            toAdd(v, u);        }        toDfs1(1, 1, 1);        toDfs2(1, 1, 1);        for (int i = 0;i < m;++i)            scanf("%d%d%d%d%I64d", &node[i].a, &node[i].b, &node[i].c, &node[i].d, &node[i].w);        sort(node, node + m);        for (int i = 0;i < m;++i)        {            divideCal(node[i].a, node[i].b, node[i].w);            divideCal(node[i].c, node[i].d, node[i].w);            toUnion(node[i].a, node[i].c, node[i].w);        }        int k = toFind(1, anc);        printf("%d %I64d\n", cnt[k], cost[k]);    }    return 0;}