Transfer water 【HDU - 4009】【最小树形图】

版权声明:https://blog.csdn.net/qq_41730082 https://blog.csdn.net/qq_41730082/article/details/85345041

题目链接

关于最小树形图的详细讲解


  挺好的一道最小树形图,然后也卡了时间,让我知道了,每次更新,缩环成点的时候,再次遍历的时候,相同的点就不要管了,即使它们之间有边,也不需要了,反正就是环内的点,已经看成了一个点,不是吗?然后解决完这个问题之后WA了几发!我的天,为什么就WA!?然后找了半天BUG,没什么问题啊,然后在看,试了下改变判断的条件,如果它们是等高的,也要+Z的话!竟然是这样,A!


#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxN = 1005;
inline ll ABS(ll x) { return x>0?x:(-x); }
int N, X, Y, Z, sum_Of_E, op;
struct node
{
    int x, y, h;
    node(int a=0, int b=0, int c=0):x(a), y(b), h(c) {}
}a[maxN];
struct Eddge
{
    int u, v;
    ll val;
    Eddge(int a=0, int b=0, ll c=0):u(a), v(b), val(c) {}
}edge[maxN*maxN*2];
ll Manhattan(int i, int j) { return abs(a[i].x - a[j].x) + abs(a[i].y - a[j].y) + abs(a[i].h - a[j].h); }
int vis[maxN], pre[maxN], id[maxN];
ll in[maxN];
ll Dir_MST(int root, int V, int E)
{
    ll ans = 0;
    while(true)
    {
        for(int i=0; i<V; i++) in[i] = INF;
        for(int i=1; i<=E; i++)
        {
            int u = edge[i].u, v = edge[i].v;
            if(u != v && in[v] > edge[i].val)   //不能是在一个环内的,浪费时间,节约时间用
            {
                in[v] = edge[i].val;
                pre[v] = u;
            }
        }
        for(int i=0; i<V; i++)
        {
            if(i == root) continue;
            if(in[i] == INF) return -1;
        }
        int cnt = 0;
        memset(vis, -1, sizeof(vis));   memset(id, -1, sizeof(id));
        in[root] = 0;
        for(int i=0; i<V; i++)
        {
            ans += in[i];
            int v = i;
            while(vis[v] != i && id[v] == -1 && v != root)
            {
                vis[v] = i;
                v = pre[v];
            }
            if(v != root && id[v] == -1)
            {
                for(int u=pre[v]; u!=v; u=pre[u]) id[u] = cnt;
                id[v] = cnt++;
            }
        }
        if(cnt == 0) return ans;
        for(int i=0; i<V; i++) if(id[i] == -1) id[i] = cnt++;
        for(int i=1; i<=E; i++)
        {
            int u = edge[i].u, v = edge[i].v;
            edge[i].u = id[u];  edge[i].v = id[v];
            if(id[u] != id[v]) edge[i].val -= in[v];
        }
        V = cnt;
        root = id[root];
    }
    return ans;
}
int main()
{
    while(scanf("%d%d%d%d", &N, &X, &Y, &Z) && (N || X || Y || Z))
    {
        sum_Of_E = 0;
        for(int i=1; i<=N; i++)
        {
            scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].h);
            edge[++sum_Of_E] = Eddge(0, i, a[i].h*X);
        }
        for(int i=1; i<=N; i++)
        {
            scanf("%d", &op);
            for(int j=1; j<=op; j++)
            {
                int e1;
                scanf("%d", &e1);
                ll tmp = 0;
                if(a[i].h < a[e1].h) tmp = Z;
                edge[++sum_Of_E] = Eddge(i, e1, Y*Manhattan(i, e1) + tmp);
            }
        }
        printf("%lld\n", Dir_MST(0, N+1, sum_Of_E));
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/qq_41730082/article/details/85345041
今日推荐