HDU 6074 Phone Call LCA + 并查集

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Problem Description

There are

dollars.

Picture from Wikimedia Commons

Input

The first line of the input contains an integer

, denoting a phone line.

Output

For each test case, print a single line containing two integers, denoting the maximum number of people that can join the party and the minimum total cost to reach that maximum number.

Sample Input

1 5 2 1 2 1 3 2 4 2 5 1 3 2 4 100 2 2 4 2 10

Sample Output

4 210

Hint

Step 1 : 1 make a phone call to 2 using line 1, the cost is 100. Step 2 : 1 make a phone call to 3 using line 1, the cost is 100. Step 3 : 2 make a phone call to 4 using line 2, the cost is 10.

题解：

　　将所有线路按代价从小到大排序，对于每条线路

　　设

　　时间复杂度

#include <bits/stdc++.h>

inline long long read(){long long x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}

using namespace std;

typedef long long LL;

const int N = 2e5 + 10, inf = 1e9;

vector<int > G[N];
struct ss{int a,b,c,d,cost;}Q[N];
int cmp(ss s1,ss s2) {return s1.cost < s2.cost;}
int sz[N],dep[N],fa[N],son[N],indexS,top[N],pos[N],ff[N],f[N];
void dfs(int u) {
    int k = 0;sz[u] = 1;dep[u] = dep[f[u]] + 1;
    for(auto to : G[u]) {
        if(to == f[u]) continue;
        f[to] = u;
        dfs(to);
        sz[u] += sz[to];
        if(sz[to] > sz[k]) k = to;
    }
    if(k) son[u] = k;
}
void dfs(int u,int chain) {
    int k = 0;pos[u] = ++indexS;
    top[u] = chain;
    if(son[u])
        dfs(son[u],chain);
    for(auto to : G[u]) {
        if(dep[to] > dep[u] && son[u] != to)
            dfs(to,to);
    }
}
int LCA(int x,int y) {
    while(top[x] != top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x,y);
        x = f[top[x]];
    }
    if(dep[x] > dep[y]) swap(x,y);
    return x;
}
LL COST[N],CNT[N];
inline int finds2(int x) {return x == fa[x] ? x:fa[x] = finds2(fa[x]);}
inline int finds(int x) {return x == ff[x] ? x:ff[x] = finds(ff[x]);}
inline void merges(int x,int y,int c) {
        int fx = finds2(x);
        int fy = finds2(y);
        if(dep[fx] < dep[fy]) swap(fx,fy);
        if(fx == fy) return;
        COST[fy] += COST[fx] + c;
        CNT[fy] += CNT[fx];
        fa[fx] = fy;
}
inline void go(int x,int zu,int c) {
    while(1) {
        x = finds(x);
        if(dep[x] <= dep[zu]) return ;
        merges(x,f[x],c);
        ff[x] = f[x];
    }
}
int n,m,T;
void init() {
    for(int i = 0; i <= n; ++i) G[i].clear(),top[i] = 0,son[i] = 0,dep[i] = 0,fa[i] = 0,pos[i] = 0;
    indexS = 0;
}
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        init();
        for(int i = 1; i < n; ++i){
            int u,v;
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        dfs(1);
        dfs(1,1);
        for(int i = 1; i <= n; ++i) fa[i] = ff[i] = i,CNT[i] = 1,COST[i] = 0;
        for(int i = 1; i <= m; ++i)
            scanf("%d%d%d%d%d",&Q[i].a,&Q[i].b,&Q[i].c,&Q[i].d,&Q[i].cost);
        sort(Q+1,Q+m+1,cmp);
        for(int i = 1; i <= m; ++i) {
            int lc = LCA(Q[i].a,Q[i].b);
            int lb = LCA(Q[i].c,Q[i].d);
            go(Q[i].a,lc,Q[i].cost);
            go(Q[i].b,lc,Q[i].cost);
            go(Q[i].c,lb,Q[i].cost);
            go(Q[i].d,lb,Q[i].cost);
            merges(lc,lb,Q[i].cost);
            //cout<<lc<<" "<<lb<<endl;
        }
        printf("%lld %lld\n",CNT[finds2(1)],COST[finds2(1)]);
    }
    return 0;
}