解題區/解題區 - UVa

UVa 10987 - Antifloyd

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

給定經由 n 個節點的無向圖，任兩點之間的最短路結果，請問原圖中的邊為何？用最少數量的邊完成。如果有可能無解則輸出 Need better measurements.

Sample Input

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7
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100
200 100
3
100
300 100

Sample Output

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7
Case #1:
2
1 2 100
2 3 100
Case #2:
Need better measurements.

Solution

最短路徑符合三角不等式，否則將會有更短的邊，檢查 f[i][j] <= f[i][k] + f[k][j]，窮舉所有可能進行檢查，$O(n^3)$ 完成。接著窮舉找到任兩個相鄰點 i, j 之間是否存在內點。若不存在內點，則表示 i, j 在原圖上有一條邊權重為 f[i][j]。

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// floyd
#include <stdio.h> 
#include <algorithm>
using namespace std;
const int MAXN = 128;
int f[MAXN][MAXN];
int anti_floyd(int n) {
    int adj[MAXN][MAXN] = {};
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            adj[i][j] = 1;
    for (int k = 0; k < n; k++) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (f[i][k] + f[k][j] < f[i][j]) {
                    puts("Need better measurements.");
                    return 0;
                }
                if (i == j || i == k || j == k)
                    continue;
                if (f[i][k] + f[k][j] == f[i][j])
                    adj[i][j] = 0;
            }
        }
    }
    
    int e = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (i == j)	continue;
            e += adj[i][j];
        }
    }
    
    printf("%d\n", e);
    for (int i = 0; i < n; i++) {
        for (int j = i+1; j < n; j++)
            if (adj[i][j])
                printf("%d %d %d\n", i+1, j+1, f[i][j]);
    }
    return 1;
}
int main() {
    int testcase, cases = 0, n;
    scanf("%d", &testcase);
    while (testcase--) {
        scanf("%d", &n);
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++)
                scanf("%d", &f[i][j]), f[j][i] = f[i][j];
            f[i][i] = 0;
        }
        
        printf("Case #%d:\n", ++cases);
        int valid = anti_floyd(n);
        puts("");
    }
    return 0;
}