Cow Tours（最短路 + Floyd）

Cow Tours

Farmer John has a number of pastures on his farm. Cow paths connect some pastures with certain other pastures, forming a field. But, at the present time, you can find at least two pastures that cannot be connected by any sequence of cow paths, thus partitioning Farmer John's farm into multiple fields.

Farmer John would like add a single a cow path between one pair of pastures using the constraints below.

A field's `diameter' is defined to be the largest distance of all the shortest walks between any pair of pastures in the field. Consider the field below with five pastures, located at the points shown, and cow paths marked by lines:

                15,15   20,15
                  D       E
                  *-------*
                  |     _/|
                  |   _/  |
                  | _/    |
                  |/      |
         *--------*-------*
         A        B       C
         10,10   15,10   20,10

The `diameter' of this field is approximately 12.07106, since the longest of the set of shortest paths between pairs of pastures is the path from A to E (which includes the point set {A,B,E}). No other pair of pastures in this field is farther apart when connected by an optimal sequence of cow paths.

Suppose another field on the same plane is connected by cow paths as follows:

                         *F 30,15
                         / 
                       _/  
                     _/    
                    /      
                   *------ 
                   G      H
                   25,10   30,10

In the scenario of just two fields on his farm, Farmer John would add a cow path between a point in each of these two fields (namely point sets {A,B,C,D,E} and {F,G,H}) so that the joined set of pastures {A,B,C,D,E,F,G,H} has the smallest possible diameter.

Note that cow paths do not connect just because they cross each other; they only connect at listed points.

The input contains the pastures, their locations, and a symmetric "adjacency" matrix that tells whether pastures are connected by cow paths. Pastures are not considered to be connected to themselves. Here's one annotated adjacency list for the pasture {A,B,C,D,E,F,G,H} as shown above:

                A B C D E F G H
              A 0 1 0 0 0 0 0 0
              B 1 0 1 1 1 0 0 0
              C 0 1 0 0 1 0 0 0
              D 0 1 0 0 1 0 0 0
              E 0 1 1 1 0 0 0 0
              F 0 0 0 0 0 0 1 0
              G 0 0 0 0 0 1 0 1
              H 0 0 0 0 0 0 1 0

Other equivalent adjacency lists might permute the rows and columns by using some order other than alphabetical to show the point connections. The input data contains no names for the points.

The input will contain at least two pastures that are not connected by any sequence of cow paths.

Find a way to connect exactly two pastures in the input with a cow path so that the new combined field has the smallest possible diameter of any possible pair of connected pastures. Output that smallest possible diameter.

PROGRAM NAME: cowtour

INPUT FORMAT

Line 1:	An integer, N (1 <= N <= 150), the number of pastures
Line 2-N+1:	Two integers, X and Y (0 <= X ,Y<= 100000), that denote that X,Y grid location of the pastures; all input pastures are unique.
Line N+2-2*N+1:	lines, each containing N digits (0 or 1) that represent the adjacency matrix as described above, where the rows' and columns' indices are in order of the points just listed.

SAMPLE INPUT (file cowtour.in)

8
10 10
15 10
20 10
15 15
20 15
30 15
25 10
30 10
01000000
10111000
01001000
01001000
01110000
00000010
00000101
00000010

OUTPUT FORMAT

The output consists of a single line with the diameter of the newly joined pastures. Print the answer to exactly six decimal places. Do not perform any special rounding on your output.

SAMPLE OUTPUT (file cowtour.out)

22.071068

 

       题意：

       给出 N（1 ~ 150），代表有 N 个坐标点（X，Y）（0 ~ 100000），后给出 N X N 的邻接矩阵，代表图的连接情况（ 0 代表不相连，1 代表相连）。图明确只有两个不相连的连通块，问如何当且仅当增加一条边，使任何点对间的最短路中的最大值最小。输出这个最小数，保留六位小数。

 

       思路：

       最短路 + Floyd。先 Floyd 预处理好所有点对之间的最短路，后枚举两个点，这两个点不相连。增加这条边后，故（ 第一个连通块任意两点间的最短路 + 增加边边长 + 第二个连通块任意两点间的最短路） 也为最短路，找到连接后所有点对最短路的最大值。每次枚举的最大值中找一个最小的，当找出这个最小值，还要与原来未连接的时候的最短路的最大值比较。最后得出的这个值即为最终结果的最大值。

 

       AC：

/*
TASK:cowtour
LANG:C++
ID:sum-g1
*/
#include <stdio.h>
#include <string.h>
#include <cmath>
#include <algorithm>
#define INF 99999999
using namespace std;

typedef struct {
    double x,y;
}node;

int n;
node no[155];
double w[155][155];
char state[155][155];
double max_dis;

double dis (double x1,double y1,double x2,double y2) {
    return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}

void floyd () {
    for (int k = 1; k <= n; ++k)
            for (int i = 1; i <= n; ++i)
                    for (int j = 1; j <= n; ++j)
                            if(w[i][k] < INF &&
                               w[k][j] < INF &&
                               w[i][j] > w[i][k] + w[k][j])
                               w[i][j] = w[i][k] + w[k][j];
}

void init () {
    for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j) {
                    if (state[i][j] == '1')
                        w[i][j] =
                        dis(no[i].x,no[i].y,no[j].x,no[j].y);
                    else w[i][j] = INF;
                    if(i == j) w[i][j] = 0;
            }
}

double make_max (int s,int e,double dis) {
    double max_now = -1;
    for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                    if(w[i][s] < INF &&
                       w[e][j] < INF)
                       max_now = max(max_now,w[i][s] + w[e][j] + dis);

    return max_now;
}

int main() {
    freopen("cowtour.in","r",stdin);
    freopen("cowtour.out","w",stdout);

    scanf("%d",&n);
    for (int i = 1; i <= n; ++i)
            scanf("%lf%lf",&no[i].x,&no[i].y);

    for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j)
                    scanf(" %c",&state[i][j]);
    }

    init();
    floyd();

    max_dis = -1;
    for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                    if(w[i][j] < INF)
                    max_dis = max(max_dis,w[i][j]);

    double min_dis = INF;
    for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                    if(w[i][j] == INF) {
                            double change = dis(no[i].x,no[i].y,no[j].x,no[j].y);
                            min_dis = min(min_dis,make_max(i,j,change));
                    }
    min_dis = max(min_dis,max_dis);

    printf("%f\n",min_dis);
    return 0;
}