P1522:牛的旅行 Cow Tours 题解

主要思路

先用 Floyed 算出初始最短路，然后再枚举两个不在同一连通块内的点进行连边并且更新当前最短路然后用 DFS 求出最长链。这道题我的垃圾做法没有\(O2\)无法 AC，时间复杂度\(O(n^4)\)。（这可能是我写过最慢的代码了）

// P1522.cpp
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#define pr pair<int, int>
#define dpr pair<double, int>
using namespace std;
const int MX_N = 200, INF = 0x3f3f3f3f;
int n;
char mat[MX_N][MX_N];
double G[MX_N][MX_N], ripe[MX_N][MX_N];
pr points[MX_N];
bool vis[MX_N];
double pow2(double num) { return num * num; }
double getDist(pr a, pr b) { return sqrt(pow2(a.first - b.first) + pow2(a.second - b.second)); }
double dfs(int u)
{
    if (vis[u])
        return 0;
    vis[u] = true;
    double res = 0;
    for (int i = 1; i <= n; i++)
        if (ripe[u][i] != INF && i != u)
            res = max(res, max(ripe[u][i], dfs(i)));
    return res;
}
int main()
{
    scanf("%d", &n);
    int x, y;
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &x, &y), points[i] = make_pair(x, y);
    for (int i = 1; i <= n; i++)
    {
        scanf("%s", mat[i] + 1);
        for (int j = 1; j <= n; j++)
            if (mat[i][j] == '1')
                G[i][j] = getDist(points[i], points[j]);
            else if (i != j)
                G[i][j] = (double)INF;
    }
    for (int k = 1; k <= n; k++)
        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]);
    double ans = (double)INF;
    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
            if (i != j && G[i][j] == INF)
            {
                memcpy(ripe, G, sizeof(G));
                memset(vis, false, sizeof(vis));
                ripe[i][j] = ripe[j][i] = getDist(points[i], points[j]);
                for (int a = 1; a <= n; a++)
                    for (int b = 1; b <= n; b++)
                        ripe[a][b] = min(ripe[a][b], ripe[a][i] + ripe[i][j] + ripe[j][b]);
                double res = dfs(i);
                ans = min(ans, res);
            }
    printf("%.6f", ans);
    return 0;
}