最短路：Bellman-Ford

题目描述

输入样例

3 3 1
1 2 1
2 3 1
1 3 3

输出样例

3

参考代码

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 510, M = 10010;

int n, m, k;
int dist[N], backup[N];     // 需要备份

struct
{
    int a, b, w;
} edges[M];

int bellman_ford()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    
    for (int i = 0; i < k; i++)
    {
        memcpy(backup, dist, sizeof dist);
        for (int j = 0; j < m; j++)
        {
            int a = edges[j].a, b = edges[j].b, w = edges[j].w;
            dist[b] = min(dist[b], backup[a] + w);
        }
    }
    
    if (dist[n] > 0x3f3f3f3f / 2) return -1;
    return dist[n];
}

int main()
{
    scanf("%d%d%d", &n, &m, &k);
    
    for (int i = 0; i < m; i++)
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        edges[i] = {a, b, w};
    }
    
    int ans = bellman_ford();
    
    printf("%d\n", ans);
    
    return 0;
}