poj 3013 Big Christmas Tree Djistra

Big Christmas Tree

题意:图中每个节点和边都有权值，图中找出一颗树，树根为1使得 Σ(树中的节点到树根的距离)*(以该节点为子树的所有节点的权值之和) 结果最小；

分析:直接求出每个节点到树根的最短距离距离，之后乘上自身节点的权值求和即可；

ps:注意特判v = 0；这时cnt = 1 > v;

Djistra + priority_queue

//Accepted    2804K    141MS
#include <cstdio>
#include <cstring>
#include <utility>
#include <queue>
#include <vector>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define inf 1LL<<40
template<typename T>
void read1(T &m)
{
    T x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>) out(a/);
    putchar(a%+'');
}
const int N = ;
typedef __int64 ll;
typedef pair<__int64,int> lli;//距离，编号
#define A first
#define B second
priority_queue<lli , vector<lli> , greater<lli> > q;
int v,val[N];
int head[N<<],tot;
struct edge{
    int to,w,Next;
}e[N<<];
void ins(int a,int b,int w = )
{
    e[++tot].Next = head[a];
    e[tot].to = b;
    e[tot].w = w;
    head[a] = tot;
}
ll d[N];bool vis[N];
ll Djistra()
{
    ll ans = ,cnt = ;
    d[] = ;
    q.push(lli{d[],});
    while(!q.empty()){
        lli t = q.top();
        q.pop();
        int u = t.B;
        if(vis[u]) continue;
        cnt++;vis[u] = true;
        ans += d[u]*val[u];
        for(int id = head[u];id;id = e[id].Next){
            int v = e[id].to,cost = e[id].w;
            if(d[v] > d[u] + cost){
                d[v] = d[u] + cost;
                q.push(lli{d[v],v});

            }
        }
    }
    if(cnt < v) return -;
    return ans;
}
int main()
{
    int T,e,kase = ;
    read1(T);
    while(T--){
        read2(v,e);
        rep1(i,,v)
            read1(val[i]),d[i] = inf,vis[i] = false;
        MS0(head);tot = ;
        rep0(i,,e){
            int a,b,w;
            read3(a,b,w);
            ins(a,b,w);ins(b,a,w);
        }
        if(v <= )out();
        else{
            while(!q.empty()) q.pop();
            ll ret = Djistra();
            if(ret == -) printf("No Answer");
            else out(ret);
        }
        puts("");
    }
    return ;
}

Djistra + heap

ps:heap中的最后的位置赋值要放在最后，不好直接在break中，因为可能是出边界了，而没有赋到需要的值；

//4116K    125MS    G++    2869B
#include <cstdio>
#include <cstring>
#include <utility>
#include <queue>
#include <vector>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define inf 1LL<<40
template<typename T>
void read1(T &m)
{
    T x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>) out(a/);
    putchar(a%+'');
}
const int N = ;
typedef __int64 ll;
typedef pair<__int64,int> lli;
#define A first
#define B second
int v,val[N];
int head[N<<],tot;
struct edge{
    int to,w,Next;
}e[N<<];
void ins(int a,int b,int w = )
{
    e[++tot].Next = head[a];
    e[tot].to = b;
    e[tot].w = w;
    head[a] = tot;
}
ll d[N];bool vis[N];
lli heap[N],q[N];int _cnt;
bool comp(lli x,lli y)
{
    return x.A < y.A;//最小堆
}
inline void push(const lli x)
{
    heap[++_cnt] = x;
    int son,fa;
    for(son = _cnt,fa = son>>;fa >= ;son = fa,fa = son>>){
        if(comp(x,heap[fa]))
            heap[son] = heap[fa];
        else    break;
    }
    heap[son] = x;
}
inline lli top()
{
    return heap[];
}
inline void pop()
{
    lli tmp =  heap[] = heap[_cnt--];
    int son ,fa;
    for(fa = ,son = fa<<;son <= _cnt;fa = son,son = fa<<){
        if(son < _cnt && comp(heap[son|],heap[son]))
            son++;
        if(comp(tmp,heap[son]))break;
        else heap[fa] = heap[son];
    }
    heap[fa] = tmp;
}
ll Djistra()
{
    ll ans = ,cnt = ;
    d[] = ;
    push(lli{d[],});
    while(_cnt){
        lli t = top();
        pop();
        int u = t.B;
        if(vis[u]) continue;
        cnt++;vis[u] = true;
        ans += d[u]*val[u];
        for(int id = head[u];id;id = e[id].Next){
            int v = e[id].to,cost = e[id].w;
            if(d[v] > d[u] + cost){
                d[v] = d[u] + cost;
                push(lli{d[v],v});
            }
        }
    }
    if(cnt < v) return -;
    return ans;
}
int main()
{
    int T,e,kase = ;
    read1(T);
    while(T--){
        read2(v,e);
        rep1(i,,v)
            read1(val[i]),d[i] = inf,vis[i] = false;
        MS0(head);tot = ;
        rep0(i,,e){
            int a,b,w;
            read3(a,b,w);
            ins(a,b,w);ins(b,a,w);
        }
        if(v <= )out();
        else{
            _cnt = ;
            ll ret = Djistra();
            if(ret == -) printf("No Answer");
            else out(ret);
        }
        puts("");
    }
    return ;
}