【树形DP】 HDU 4756 Install Air Conditioning

通道

题意：给n个点，现在要使这n个点连通，并且要求代价最小。现在有2个点之间不能直接连通（除了第一个点），求最小代价

思路:先求mst，然后枚举边，对于生成树上的边替换，用树形dp O(N^2)求出每条生成树边的最小替代边。然后替换后的最大值

代码：

#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int inf = 0x3f3f3f3f;

inline void RD(int &ret) {
    char c;
    int flag = 1 ;
    do {
        c = getchar();
        if(c == '-')flag = -1 ;
    } while(c < '0' || c > '9') ;
    ret = c - '0';
    while((c=getchar()) >= '0' && c <= '9')
        ret = ret * 10 + ( c - '0' );
    ret *= flag ;
}
struct kdq{
    int s , e  ;
    double l ;
    bool operator <  (const kdq & fk)const {
        return l > fk.l ;
    }
} ;

#define N 1111
#define M 111111
int st[M] ;
int top ;
double NMIN[N][N] ;
int x[N] , y[N] , cost , n ;
double Map[N][N] ;
double dis[N]  ;
bool vis[N] ;
vector<int>G[N] ;
priority_queue<kdq>qe ;
double mst = 0 ;
bool ismst[N][N] ;
double MINE[N][N] ;
double getdis(int i , int j){
    return sqrt(1.0 * (x[i] - x[j]) * (x[i] - x[j]) + 1.0 * (y[i] - y[j]) * (y[i] - y[j])) ;
}


void Prim(){
    mst = 0 ;
    memset(ismst, 0, sizeof ismst);
    while(!qe.empty())qe.pop() ;
    for (int i = 0 ; i < n ; i ++ )dis[i] = Map[0][i] , vis[i] = 0 ,G[i].clear() ,qe.push((kdq){0 , i , dis[i]}) ;
    dis[0] = 0 , vis[0] = 1 ;
    while(!qe.empty()){
        kdq tp = qe.top() ; qe.pop() ;
        if(vis[tp.e])continue ;
        mst += Map[tp.s][tp.e] ;
        vis[tp.e] = 1 ;
        G[tp.s].push_back(tp.e) ; G[tp.e].push_back(tp.s) ;
        ismst[tp.e][tp.s] = ismst[tp.s][tp.e] = 1 ;
        MINE[tp.s][tp.e] = MINE[tp.e][tp.s] = inf ;//删除树边
        for (int i = 0 ; i < n ; i ++ )if(!vis[i] && dis[i] > Map[tp.e][i])dis[i] = Map[tp.e][i] , qe.push((kdq){tp.e , i , dis[i]}) ;
    }
}

void dfs(int root , int fa , int now){
    int sz = G[now].size() ;
    for (int i = 0 ; i < sz ; i ++ ){
        int e = G[now][i] ;
        if(e != fa)dfs(root , now , e) , MINE[root][now] = min(MINE[root][now] , MINE[root][e]) ;
    }
}

void dfs(int fa, int now){
    int fk = top ;
    st[ ++ top] = now ;int sz = G[now].size() ;
    for (int i = 0 ; i < sz ; i ++ ){
        int e = G[now][i] ; if(e != fa)dfs(now , e) ;
    }
    if(fa != -1){
        NMIN[now][fa] = NMIN[fa][now] = inf ;
        for (int i = fk + 1 ; i <= top ; i ++ )NMIN[now][fa] = NMIN[fa][now] = min(NMIN[fa][now] , MINE[st[i]][fa]) ;
    }
}
void solve(){scanf("%d%d", &n, &cost);
    for (int i = 0 ; i < n ; i ++ )RD(x[i]) , RD(y[i]) ;
    for (int i = 0 ; i < n ; i ++ )
        for (int j = 0 ; j < n ; j ++ )
            MINE[i][j] = Map[i][j] = (i == j) ? 0 : getdis(i , j) ;
    Prim() ;
    for (int i = 0 ; i < n ; i ++ )dfs(i , -1 , i ) ;
    top = 0 ;
    dfs(-1 , 0) ;
    double ans = 0 ;
    for (int i = 0 ; i < n ; i ++ ){
        for (int j = 0 ; j < n ; j ++ ){
            if(i == 0 || j == 0)continue ;
            if(!ismst[i][j])continue ;
            ans = max(ans , NMIN[i][j] - Map[i][j]) ;
        }
    }
    printf("%.2f\n",(ans + mst) * cost) ;
}
int main() {
    int _ ; scanf("%d", &_); while(_ --)solve() ;
    return 0 ;
}