最小环(floyd以及dijkstra实现+例题）
最小环定义
最小环是指在一个图中，有n个节点构成的边权和最小的环(n>=3)。
一般来说,最小环分为有向图最小环和无向图最小环。
最小环算法：
直接暴力：
设\(u\)和\(v\)之间有一条边长为\(w\)的边,\(dis(u,v)\)表示删除\(u\)和\(v\)之间的连边之后,\(u\)和\(v\)之间的最短路。那么最小环是\(dis(u,v)+w\)总时间复杂度\(O(n^2m)\)。
Dijkstra
任意一个环，假设连接\(u\)和\(v\)，我们都可以看做删除\(u\)与\(v\)的直接连边之后的\(u\)到\(v\)的最短路再加上该边，若边数为\(m\)，那么找出最小环要跑\(m\)次Dijkstra，复杂度为\(O(n(n+m)log)\)
Floyd
记原图\(u\),\(v\)之间边权为\(mp(u,v)\),floyd算法在外层循环到第k个点时(还没开始第k次循环),最短路数组\(g\)中，\(g(u,v)\)表示的是从\(u\)到\(v\)且仅经过编号\([1,k)\)区间中的点的最短路。
最小环至少有三个顶点，设其中编号最大的顶点编号为\(w\)，环上与\(w\)相邻两侧的两个点为\(u\),\(v\),则在最外层循环枚举到\(k=w\)时，该环的长度为\(g(u,v)+mp(v,w)+mp(w,u)\)，所以在循环时候\(i\),\(j\)只需枚举到\(i<k\),\(j<k\)，更新答案即可
复杂度：\(O(n^3)\)
下面是实现参考:
for (int k = 1; k <= n; k++)
{
for (int i = 1; i < k; i++)
for (int j = i + 1; j < k; j++)
ans = min(ans, g[i][j] + mp[i][k] + mp[k][j]);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
}
例题 HDU 1599
此处输入链接的描述
裸题= =，没啥东西,就去个重
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
//clock_t c1 = clock();
//std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e2 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
ll g[MAXN][MAXN];
ll mp[MAXN][MAXN]; //原图
int n, m;
void init()
{
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
mp[i][j] = i == j ? 0 : inf;
}
int main()
{
while (~scanf("%d%d", &n, &m))
{
init();
memset(g, inf, sizeof(g));
for (int i = 0; i < m; i++)
{
int u, v;
ll val;
scanf("%d%d%lld", &u, &v, &val);
if (val < mp[u][v])
mp[v][u] = mp[u][v] = val;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
g[i][j] = mp[i][j];
ll ans = inf;
for (int k = 1; k <= n; k++)
{
for (int i = 1; i < k; i++)
for (int j = i + 1; j < k; j++)
ans = min(ans, g[i][j] + mp[i][k] + mp[k][j]);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
}
if (ans == inf)
printf("It's impossible.\n");
else
printf("%d\n", ans);
}
return 0;
}
参考oiwiki
dijkstra版拓展题 HDU-6005 Pandaland
解题思路：
也是最小环（当然啦啊喂），不过点是坐标的形式，需要转换为序号，之后记录跑的起点和终点，dijkstra中不跑这条边即可，可以稍微剪剪枝（超重要诶），如果dijkstra中当前最短边和直连边权值之和大于ans,直接break,会快很多（迫真）,这题没重边，所以不用去
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
//clock_t c1 = clock();
//std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 8e3 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
struct node
{
int v, c;
node(int a = 0, int b = 0) { v = a, c = b; }
bool operator<(const node &a) const
{
if (c == a.c)
return v < a.v;
else
return c > a.c;
}
};
struct Edge
{
int v, cost;
Edge(int _v = 0, int _cost = 0) { v = _v, cost = _cost; }
};
struct EE
{
int u, v, w;
EE(int _u = 0, int _v = 0, int _w = 0) { u = _u, v = _v, w = _w; }
} E[MAXN];
vector<Edge> G[MAXN];
bool vis[MAXN];
int dist[MAXN];
//点的编号从1开始
int ans;
void Dijkstra(int n, int start, int ed, int val)
{
memset(vis, false, sizeof(vis));
for (int i = 1; i <= n; i++)
dist[i] = inf;
priority_queue<node> que;
while (!que.empty())
que.pop();
dist[start] = 0;
que.push(node(start, 0));
node temp;
while (!que.empty())
{
temp = que.top();
que.pop();
int u = temp.v;
if (temp.c + val > ans)
break; //该点最短距离+start与ed的直连边长大于ans,说明不是最小环,直接break
if (vis[u])
continue;
vis[u] = true;
for (int i = 0; i < G[u].size(); i++)
{
int v = G[u][i].v;
int cost = G[u][i].cost;
if ((u == start && v == ed) || (v == ed && u == start)) //不经过start-ed的直达边
continue;
if (!vis[v] && dist[v] > dist[u] + cost)
{
dist[v] = dist[u] + cost;
que.push(node(v, dist[v]));
}
}
}
}
void addedge(int u, int v, int w)
{
G[u].push_back(Edge(v, w));
}
int cnt = 0;
void init(int n)
{
for (int i = 1; i <= n; i++)
G[i].clear();
cnt=0;
}
/* 题目没重边，不然还得去一波重 */
int main()
{
int t;
scanf("%d", &t);
for (int tt = 1; tt <= t; tt++)
{
map<PII, int> mp;
int m;
scanf("%d", &m);
init(m << 1);
for (int i = 0; i < m; i++)
{
int x1, y1, x2, y2, w;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &w);
if (!mp[PII(x1, y1)])
mp[PII(x1, y1)] = ++cnt;
if (!mp[PII(x2, y2)])
mp[PII(x2, y2)] = ++cnt;
int u = mp[PII(x1, y1)], v = mp[PII(x2, y2)];
addedge(u, v, w);
addedge(v, u, w);
E[i] = EE(u, v, w); //存原边
}
ans = inf;
for (int i = 0; i < m; i++)
{
if (ans < E[i].w)
continue;
Dijkstra(cnt, E[i].u, E[i].v, E[i].w);
ans = min(ans, dist[E[i].v] + E[i].w);
}
printf("Case #%d: %d\n", tt, ans == inf ? 0 : ans);
}
return 0;
}
