P5764 [CQOI2005]新年好
关于spfa他已经死了 …
思路:
对于必须到达的点,我们可以利用深搜,搜索出其排列的所有情况,其次再以每一个必达点为源点进行最短路操作分别记录下其路径距离,再依次对每种排列进行查找。
但是上述方法会使得时间复杂度过高,所以,我们只需要将两步操作交换顺序,先对每一个必达点进行最短路操作,再利用dfs来找出路径最短值即可。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<cstdio>
#include<cmath>
#include<iomanip>
#include<sstream>
#include<queue>
#define pi 3.1415926535
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
typedef pair<int, int> pii;
const int N = 1e6 + 10;
int h[N], e[N], ne[N], idx, w[N];
void add(int a, int b, int c) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
int n, m;
int id[N];
int dis[6][N];
bool st[N];
void dijkstra(int x, int dis[]) {
memset(dis, 0x3f, N * 4);
memset(st, false, sizeof st);
dis[x] = 0;
priority_queue<pii, vector<pii>, greater<pii>> heap;
heap.push({ 0,x });
while (heap.size())
{
pii t = heap.top();
heap.pop();
int ver = t.second;
if (st[ver])continue;
st[ver] = true;
for (int i = h[ver]; i != -1; i = ne[i]) {
int j = e[i];
if (dis[j] > dis[ver] + w[i]) {
dis[j] = dis[ver] + w[i];
heap.push({ dis[j],j });
}
}
}
}
int dfs(int num, int start, int dist) {
if (num > 5) return dist;
int res = 0x3f3f3f3f;
for(int i = 1; i <= 5; i++) {
if (!st[i]) {
st[i] = true;
res = min(res, dfs(num + 1, i, dist + dis[start][id[i]]));
st[i] = false;
}
}
return res;
}
int main() {
cin >> n >> m;
id[0] = 1;
memset(h, -1, sizeof h);
for (int i = 1; i <= 5; i++)cin >> id[i];
while (m--)
{
int a, b, c;
cin >> a >> b >> c;
add(a, b, c);
add(b, a, c);
}
for (int i = 0; i < 6; i++)dijkstra(id[i], dis[i]);
memset(st, false, sizeof st);
cout << dfs(1, 0, 0) << endl;
}