洛谷 - P1522 - 牛的旅行 - Cow Tours - Floyd

https://www.luogu.org/problem/P1522

好坑啊，居然还有直径不通过新边的数据，还好不是很多。

注意一定要等Floyd跑完之后再去找连通块的直径，不然一定是INF。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
int pre[155];

void init(int n) {
    for(int i = 0; i < n; ++i)
        pre[i] = i;
}

double D[155];

int find(int x) {
    return pre[x] == x ? x : pre[x] = find(pre[x]);
}
void unit(int x, int y) {
    int fx = find(x), fy = find(y);
    if(!(fx == fy))
        pre[fx] = fy;
}
bool iscc(int x, int y) {
    return find(x) == find(y);
}
double pos[155][2];
double dis_t[155][155];
double dist(int i, int j) {
    return sqrt((pos[i][0] - pos[j][0]) * (pos[i][0] - pos[j][0]) + (pos[i][1] - pos[j][1]) * (pos[i][1] - pos[j][1]));
}
int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    init(150);
    int n;
    cin >> n;
    for(int i = 0; i < n; ++i)
        cin >> pos[i][0] >> pos[i][1];
    char ch;
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j)
            dis_t[i][j] = 1e9;
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j) {
            cin >> ch;
            if(ch == '1') {
                unit(i, j);
                dis_t[j][i] = dis_t[i][j] = dist(i, j);
            }
            if(i == j)
                dis_t[i][j] = 0;
        }
    for(int k = 0; k < n; ++k)
        for(int j = 0; j < n; ++j)
            for(int i = 0; i < n; ++i) {
                if(iscc(i, j)) {
                    dis_t[i][j] = min(dis_t[i][j], dis_t[i][k] + dis_t[k][j]);
                }
            }
    double max_dis[155] = {};
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            if(iscc(i, j)){
                max_dis[i] = max(max_dis[i], dis_t[i][j]);

            }
        }
        D[find(i)] = max(D[find(i)], max_dis[i]);
    }
    double minn = 1e9;
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j)
            if(!iscc(i, j)) {
                minn = min(minn, max(max(D[find(i)], D[find(j)]), dist(i, j) + max_dis[i] + max_dis[j]));
            }
    /*char s[20005];
    sprintf(s,"%.8f\n", minn);
    int pi=0;
    for(pi=0;s[pi]!='.';++pi);
    s[pi+7]='\0';
    puts(s);*/
    printf("%.6f\n", minn);
    return 0;
}