# USACO 2.4 Cow Tours 牛的旅行 解题报告

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 

John想在农场里添加一条路径(注意，恰好一条)。对这条路径有以下限制：

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

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

 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 

PROGRAM NAME: cowtour
INPUT FORMAT:
(file cowtour.in)

OUTPUT FORMAT:
(file cowtour.out)

SAMPLE INPUT
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
SAMPLE OUTPUT
22.071068
====================== 华丽的分割线 ======================

point是一个结构体, point[i] 表示第i个牧区的坐标:

struct point{
int x, y;
}point[MAX];

C语言:

/*
LANG: C
ID: zqy11001
PROG: cowtour
*/
#include <stdio.h>
#define INF (1e5)
#define MAX (150)
#define getint(i) scanf(%d, &i)

struct point{
int x, y;
}point[MAX];
double dis[MAX][MAX];
double diam[MAX];
double fdiam[MAX];
int fie[MAX];
int n;

double getdis(int i, int j)
{
struct point *a, *b;
a = &point[i];
b = &point[j];
return sqrt((double)(a->x - b->x)*
(a->x - b->x) +
(double)(a->y - b->y)*
(a->y - b->y));
}

void mark(int i, int m)
{
int j;
if(fie[i] != 0){
return ;
}
fie[i] = m;
for(j = 0; j < n; j++){
if(dis[i][j] < INF){
mark(j, m);
}
}
}

int main(void)
{
int i, j, k;
int c, now = 1;
double t, max;
freopen(cowtour.in, r, stdin);
freopen(cowtour.out, w, stdout);
getint(n);
for(i = 0; i < n; i++){
getint(point[i].x);
getint(point[i].y);
}

for(i = 0; i < n; i++){
getchar();
for(j = 0; j < n; j++){
c = getchar();
if(i == j){
dis[i][j] = 0;
}else if(c == \'0\'){
dis[i][j] = INF;
}else{
dis[i][j] = getdis(i, j);
}
}
}

for(i = 0; i < n; i++){
if(fie[i] == 0){
mark(i, now++);
}
}

for(k = 0; k < n; k++)
for(i = 0; i < n; i++)
for(j = 0; j < n; j++){
if(dis[i][j] > dis[i][k] + dis[k][j]){
dis[i][j] = dis[i][k] + dis[k][j];
}
}

for(i = 0; i < n; i++){
for(j = 0; j < n; j++){
if(diam[i] < dis[i][j] && dis[i][j] < INF){
diam[i] = dis[i][j];
}
}
if(fdiam[fie[i]] < diam[i]){
fdiam[fie[i]] = diam[i];
}
}

max = INF;
for(i = 0; i < n; i++){
for(j = 0; j < n; j++){
if(fie[i] == fie[j]){
continue;
}

t = diam[i] + diam[j] + getdis(i, j);
if(t < fdiam[fie[i]]){
t = fdiam[fie[i]];
}
if(t < fdiam[fie[j]]){
t = fdiam[fie[j]];
}

if(t < max){
max = t;
}
}
}
printf(%.6lf\\n, max);
return 0;
}`