描述
题解
最小树形图,模版题,之所以这么说有两个原因,第一,真的是模版,第二,我只会套模版,增加超级源点搞搞。
如果有比较好的最小树形图的算法资料,欢迎推荐~~~
代码
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
using namespace std;
/* * 最小树形图 * int型 * 复杂度O(NM) * 点从0开始 */
const int INF = 0x3f3f3f3f;
const int MAXN = 1010;
const int MAXM = 1000010;
struct Edge
{
int u, v, cost;
};
Edge edge[MAXM];
int pre[MAXN], id[MAXN], visit[MAXN], in[MAXN];
int zhuliu(int root, int n, int m)
{
int res = 0, v;
while (1)
{
memset(in, 0x3f, sizeof(in));
for (int i = 0; i < m; i++)
{
if (edge[i].u != edge[i].v && edge[i].cost < in[edge[i].v])
{
pre[edge[i].v] = edge[i].u;
in[edge[i].v] = edge[i].cost;
}
}
for (int i = 0; i < n; i++)
{
if (i != root && in[i] == INF)
{
return -1; // 不存在最小树形图
}
}
int tn = 0;
memset(id, -1, sizeof(id));
memset(visit, -1, sizeof(visit));
in[root] = 0;
for (int i = 0; i < n; i++)
{
res += in[i];
v = i;
while (visit[v] != i && id[v] == -1 && v != root)
{
visit[v] = i;
v = pre[v];
}
if (v != root && id[v] == -1)
{
for (int u = pre[v]; u != v ; u = pre[u])
{
id[u] = tn;
}
id[v] = tn++;
}
}
if (tn == 0)
{
break; // 没有有向环
}
for (int i = 0; i < n; i++)
{
if (id[i] == -1)
{
id[i] = tn++;
}
}
for (int i = 0; i < m; i++)
{
v = edge[i].v;
edge[i].u = id[edge[i].u];
edge[i].v = id[edge[i].v];
if (edge[i].u != edge[i].v)
{
edge[i].cost -= in[v];
}
}
n = tn;
root = id[root];
}
return res;
}
// 结点坐标
struct Node
{
int x;
int y;
int z;
} node[MAXN];
int getDis(Node a, Node b)
{
return abs(a.x - b.x) + abs(a.y - b.y) + abs(a.z - b.z);
}
int main()
{
int n, X, Y, Z;
int num, key;
while (cin >> n >> X >> Y >> Z, n + X + Y + Z)
{
for (int i = 0; i < n; i++)
{
scanf("%d%d%d", &node[i].x, &node[i].y, &node[i].z);
}
int pos = 0;
for (int i = 0; i < n; i++)
{
scanf("%d", &num);
while (num--)
{
scanf("%d", &key);
edge[pos].cost = getDis(node[i], node[--key]) * Y;
if (node[key].z > node[i].z)
{
edge[pos].cost += Z;
}
edge[pos].u = i;
edge[pos++].v = key;
}
}
for (int i = 0; i < n; i++)
{
edge[pos].u = n;
edge[pos].v = i;
edge[pos++].cost = node[i].z * X;
}
printf("%d\n", zhuliu(n, n + 1, pos));
}
return 0;
}