思路
从自己家开始,顺序任意,能去五个亲戚家,可以从亲戚家去到另外的亲戚家,于是这启发我们把每个亲戚和自己到全图其他点的最短路处理出来。
这乍一看是多源汇最短路,但是我们发现\(Floyd\)算法是\(O(N^{3})\)的,在这题的条件下=根本跑不过
但是我们的源点有几个? 只有一个自己加上五个亲戚,如果做\(6\)次\(Dijkstra\),每次\(Dijkstra\)是\(O(mlogn)\)的,所以我们可以用\(Dijkstra\)
代码
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 1000010, M = 1e5 + 10, INF = 0x3f3f3f3f;
int h[N], e[M << 1], w[M << 1], ne[M << 1], idx;
int dist[6][N], rel[10], n, m;
bool st[N];
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}
void Dijkstra(int S, int dist[])
{
for (int i = 0; i <= n; i ++ )
dist[i] = INF;
memset(st, 0, sizeof st);
priority_queue<PII, vector<PII>, greater<> > heap;
heap.push({0, S}), dist[S] = 0;
while (heap.size())
{
auto t = heap.top().second; heap.pop();
if (st[t]) continue ;
st[t] = true;
for (int i = h[t]; ~i; i = ne[i])
{
int j = e[i];
if (dist[j] > dist[t] + w[i])
{
dist[j] = dist[t] + w[i];
heap.push({dist[j], j});
}
}
}
}
int dfs(int cur, int S, int dis) //cur是已经访问过的亲戚,S是当前所在点,dis是这条路线的总长度
{
if (cur == 6) return dis;
int res = INF;
for (int i = 1; i <= 5; i ++ )
{
if (st[i]) continue ;
st[i] = true;
int nx = rel[i];
res = min(res, dfs(cur + 1, i, dist[S][nx] + dis));
st[i] = false;
}
return res;
}
int main()
{
memset(h, -1, sizeof h);
scanf("%d%d", &n, &m);
rel[0] = 1;
for (int i = 1; i <= 5; i ++ )
scanf("%d", &rel[i]);
while (m -- )
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
add(u, v, w), add(v, u, w);
}
for (int i = 0; i <= 5; i ++ ) Dijkstra(rel[i], dist[i]);
memset(st, 0, sizeof st);
printf("%d\n", dfs(1, 0, 0));
return 0;
}
标签:dist,cur,int,1135,新年好,heap,include,dis
From: https://www.cnblogs.com/StkOvflow/p/16994922.html