luogu1522 牛的旅行

题目大意

　　每个牧场里的某些坐标位置有牧区，牧区间有一个个路径（长度为位置间的直线距离）。一个连通块内两个节点间的最短路径长度最大值为它的直径。请编程找出一条连接两个不同牧场的路径，使得连上这条路径后，这个更大的新牧场有最小的直径。输出在所有牧场中最小的可能的直径。

思路

注意

　　Floyd先枚举k。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <cassert>
using namespace std;


const int MAX_NODE = 200;
const double INF = 200000;
double Dist[MAX_NODE][MAX_NODE];
int TotNode, TotBlock;

struct Coor//Coordinate
{
    int X, Y;

    double Dist(const Coor& a)const
    {
        double x1 = X, y1 = Y, x2 = a.X, y2 = a.Y;
        double dx = abs(x1 - x2), dy = abs(y1 - y2);
        return sqrt(dx * dx + dy * dy);
    }
};

struct Node
{
    Coor Coordinate;
    double MaxDist;
}_nodes[MAX_NODE];

void InitDist()
{
    for (int i = 0; i < MAX_NODE; i++)
        for (int j = 0; j < MAX_NODE; j++)
            Dist[i][j] = INF;
    for (int i = 0; i < MAX_NODE; i++)
        Dist[i][i] = 0;
}

void Read()
{
    scanf("%d", &TotNode);
    for (int i = 1; i <= TotNode; i++)
        scanf("%d%d
", &_nodes[i].Coordinate.X, &_nodes[i].Coordinate.Y);
    char s[MAX_NODE];
    for (int i = 1; i <= TotNode; i++)
    {
        scanf("%s", s + 1);
        for (int j = 1; j <= TotNode; j++)
            if (s[j] - '0')
                Dist[i][j] = _nodes[i].Coordinate.Dist(_nodes[j].Coordinate);
    }
}

void Floyd()
{
    for (int k = 1; k <= TotNode; k++)
        for (int i = 1; i <= TotNode; i++)
            for (int j = 1; j <= TotNode; j++)
                Dist[i][j] = min(Dist[i][j], Dist[i][k] + Dist[k][j]);
}

double MaxMaxDist;
void GetNodeMaxDist()
{
    for (int i = 1; i <= TotNode; i++)
        for (int j = 1; j <= TotNode; j++)
            if (Dist[i][j] < INF)
            {
                _nodes[i].MaxDist = max(_nodes[i].MaxDist, Dist[i][j]);
                MaxMaxDist = max(MaxMaxDist, _nodes[i].MaxDist);
            }
}

double GetAns()
{
    double ans = INF;
    for (int i = 1; i <= TotNode; i++)
        for (int j = 1; j <= TotNode; j++)
            if (Dist[i][j] == INF)
                ans = min(ans, _nodes[i].MaxDist + _nodes[j].MaxDist + _nodes[i].Coordinate.Dist(_nodes[j].Coordinate));
    return max(ans, MaxMaxDist);
}

int main()
{
    InitDist();
    Read();
    Floyd();
    GetNodeMaxDist();
    printf("%.6f
", GetAns());
    return 0;
}