https://cn.vjudge.net/problem/HDU-4085
題意:
給你 n 房子, m 條路, k 戶人家,每條路有一個權值,表示修復這條邊需要的代價,起初 k 戶人家依次住在第 1個,第 2 個,,,第 k 個房子裏,現要求讓着前 k 個房子和後
k 個房子一一對應連通建路,問最小的代價是多少。
分析:
可以先簡單看成一個斯坦納樹問題:n 個點選給定的 2 * k 個點的最小生成樹。但注意題意要求一一對應,所以最後還需要 dp2[],這很好理解,舉個例子,四個點的最小
生成樹權和不一定等於分割開的兩棵最小生成樹的權和。
dp[i][j] 表示以 j 爲起點的包含點的狀態爲 i 的最小生成樹。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <bitset>
typedef long long int ll;
const int MOD = (int)1e9 + 7;
const int INF = 99999999;
using namespace std;
struct Edge
{
int v;
int w;
int next;
};
int t;
int n, m, k;
int cnt;
int top;
int s[55];
int vis[1 << 10][55];
int dp1[1 << 10][55];
int dp2[1 << 10];
Edge edge[2005];
int head[55];
queue<pair<int, int> > Q;
void add_edge(int u, int v, int w)
{
edge[cnt].v = v;
edge[cnt].w = w;
edge[cnt].next = head[u];
head[u] = cnt++;
edge[cnt].v = u;
edge[cnt].w = w;
edge[cnt].next = head[v];
head[v] = cnt++;
}
bool check(int s)
{
int res = 0;
for (int i = 0; s != 0; i++)
{
if (s & 1)
res += (i < k) ? 1 : -1;
s >>= 1;
}
return (res == 0);
}
void spfa()
{
while (!Q.empty())
{
int state = Q.front().first;
int u = Q.front().second;
Q.pop();
vis[state][u] = 0;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v;
int w = edge[i].w;
if (dp1[state | s[v]][v] > dp1[state][u] + w)
{
dp1[state | s[v]][v] = dp1[state][u] + w;
if ((state | s[v]) == state && !vis[state][v])
{
Q.push(make_pair(state, v));
vis[state][v] = 1;
}
}
}
}
}
int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%d %d %d", &n, &m, &k);
cnt = 0;
top = 1 << (2 * k);
memset(s, 0, sizeof(s));
memset(vis, 0, sizeof(vis));
memset(head, -1, sizeof(head));
for (int i = 0; i < top; i++)
for (int j = 0; j < n; j++)
dp1[i][j] = INF;
int u, v, w;
for (int i = 0; i < m; i++)
{
scanf("%d %d %d", &u, &v, &w);
add_edge(u - 1, v - 1, w);
}
for (int i = 0; i < k; i++)
{
s[i] = 1 << i;
dp1[s[i]][i] = 0;
s[n - 1 - i] = 1 << (k + i);
dp1[s[n - 1 - i]][n - 1 - i] = 0;
}
for (int i = 0; i < top; i++)
{
for (int j = 0; j < n; j++)
{
for (int t = (i - 1) & i; t != 0; t = (t - 1) & i)
dp1[i][j] = min(dp1[i][j], dp1[t | s[j]][j] + dp1[(i - t) | s[j]][j]);
if (dp1[i][j] < INF)
{
Q.push(make_pair(i, j));
vis[i][j] = 1;
}
}
spfa();
}
for (int i = 0; i < top; i++)
{
dp2[i] = INF;
for (int j = 0; j < n; j++)
{
if (i & (1 << j))
{
dp2[i] = min(dp2[i], dp1[i][j]);
break;
}
}
}
for (int i = 0; i < top; i++)
{
if (check(i))
{
for (int t = (i - 1) & i; t != 0; t = (t - 1) & i)
{
if (check(t))
dp2[i] = min(dp2[i], dp2[t] + dp2[i - t]);
}
}
}
if (dp2[top - 1] >= INF)
printf("No solution\n");
else
printf("%d\n", dp2[top - 1]);
}
return 0;
}