数据结构 娱乐中心选址 (Floyd+暴力)

Description

有n个地点编号1～n，在其中一个地点建立娱乐中心，要求该点距其它各地点的最长往返路程最短，相同条件下总的往返路程和越短越好，如果仍相同，取编号最小的地点，问娱乐中心应选址何处？

Input

第一行输入测试数据组数。每组数据第一行输入地点数n（2≤n≤300），路径数m（0≤m≤10000）。接下来m行，每行有一条有向边，输入起终点u、v（1≤u,v≤n），路径长度d（0≤d≤30000）。

Output

对每一个测试样例输出两行。第一行输出测试样例组数。第二行，若能找到可以到达所有地点的选址，输出最佳选址点的编号、最长往返路程、往返路程和；否则，输出-1。每两组测试数据输出一个空行。

Sample Input

2

3 4

1 2 1

2 1 2

2 3 10

3 123

2 1

1 2 1

Sample Output

Case #1:

1 34 37

 

Case #2:

-1

HINT

考察知识点：图的Floyd算法，时间复杂度O(n^3)，空间复杂度O(m*n)

Append Code

析：由于 n 比较小，我们先用Floyd把任意两个结点的最短路长度示求出来，然后再遍历所有任意两个结点，去枚举哪个点做娱乐地址，求出最优的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 300 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn][maxn];
 
int main(){
    int T;  scanf("%d", &T);
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                dp[i][j] = i == j ? 0 : INF;
        int u, v, d;
        for(int i = 0; i < m; ++i){
            scanf("%d %d %d", &u, &v, &d);
            if(d < dp[u][v])  dp[u][v] = d;
        }
 
        for(int k = 1; k <= n; ++k)
            for(int i = 1; i <= n; ++i)
                for(int j = 1; j <= n; ++j)
                        if(dp[i][j] > dp[i][k] + dp[k][j]) dp[i][j] = dp[i][k] + dp[k][j];
        int ans = -1, ans1 = INF, ans2 = INF;
        for(int i = 1; i <= n; ++i){
            int sum1 = 0, sum2 = 0;
            for(int j = 1; j <= n; ++j){
                if(sum1 > ans1) break;
                int tmp = dp[i][j] + dp[j][i];
                if(sum1 < tmp)  sum1 = tmp;
                sum2 += tmp;
            }
            if(sum1 < ans1 || (sum1 == ans1 && sum2 < ans2)){
                ans1 = sum1;
                ans2 = sum2;
                ans = i;
            }
        }
 
        if(kase != 1)  printf("
");
        printf("Case #%d:
", kase);
        if(ans == -1)  printf("-1
");
        else printf("%d %d %d
", ans, ans1, ans2);
    }
    return 0;
}