带负边权的最短路径（有向图）——Bellman-Ford算法

参考了一位大佬的代码，一直很喜欢简洁的代码。（来源及出处已附上）

#include <iostream>
#include <cstring>
#include <cstdio>

#define MAX 100
#define INF 0x3f3f3f
using namespace std;
//有向图
struct Edge
{
    int u,v,cost;
}e[MAX];
int dist[MAX];  //最短路径
int prev[MAX];  //路径
int m,n;    //边数和顶点数

bool Bellman_Ford(int v0)
{
    int u=v0;
    for(int i=1;i<=n;i++)
        dist[i]=INF;
    dist[u]=0;
    for(int i=1;i<=n;i++)
        for(int j=0;j<m;j++)
            if(dist[e[j].v]>dist[e[j].u]+e[j].cost)
            {
                dist[e[j].v]=dist[e[j].u]+e[j].cost;
                prev[e[j].v]=e[j].u;
            }
    for(int i=0;i<m;i++)
        if(dist[e[i].v]>dist[e[i].u]+e[i].cost)
            return 0;
    return 1;
}

int main()
{
    cin>>n>>m;
    for(int i=0;i<m;i++)
        cin>>e[i].u>>e[i].v>>e[i].cost;
    if(Bellman_Ford(1))
        for(int i = 1; i <= n; ++i) //每个点最短路
        {
            printf("%d
", dist[i]);
        }
    else
        printf("have negative circle
");
    return 0;
}
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版权声明：本文为CSDN博主「weixin_43249938」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/weixin_43249938/article/details/88217729

例题：

The Preliminary Contest for ICPC Asia Nanjing 2019  H. Holy Grail

As the current heir of a wizarding family with a long history,unfortunately, you find yourself forced to participate in the cruel Holy Grail War which has a reincarnation of sixty years.However,fortunately,you summoned a Caster Servant with a powerful Noble Phantasm.When your servant launch her Noble Phantasm,it will construct a magic field,which is actually a directed graph consisting of n vertices and m edges.More specifically,the graph satisfies the following restrictions :

Does not have multiple edges(for each pair of vertices x and y, there is at most one edge between this pair of vertices in the graph) and does not have self-loops(edges connecting the vertex with itself).
 May have negative-weighted edges.
 Does not have a negative-weighted loop.
 n<=300 , m<=500.
Currently,as your servant's Master,as long as you add extra 6 edges to the graph,you will beat the other 6 masters to win the Holy Grail.

However，you are subject to the following restrictions when you add the edges to the graph:

Each time you add an edge whose cost is c,it will cost you c units of Magic Value.Therefore,you need to add an edge which has the lowest weight(it's probably that you need to add an edge which has a negative weight).
 Each time you add an edge to the graph,the graph must not have negative loops,otherwise you will be engulfed by the Holy Grail you summon.

Input

Input data contains multiple test cases. The first line of input contains integer t — the number of testcases (1≤t≤5).

For each test case,the first line contains two integers n,m,the number of vertices in the graph, the initial number of edges in the graph.

Then m lines follow, each line contains three integers x, y and w (0≤x,y<n,-10^9≤w≤10^9, x�​=y) denoting an edge from vertices x to y (0-indexed) of weight w.

Then 6 lines follow, each line contains two integers s,t denoting the starting vertex and the ending vertex of the edge you need to add to the graph.

It is guaranteed that there is not an edge starting from s to t before you add any edges and there must exists such an edge which has the lowest weight and satisfies the above restrictions, meaning the solution absolutely exists for each query.

 

Output

For each test case,output 666 lines.

Each line contains the weight of the edge you add to the graph.

样例输入：
1
10 15
4 7 10
7 6 3
5 3 3
1 4 11
0 6 20
9 8 25
3 0 9
1 2 15
9 0 27
5 2 0
7 3 -5
1 7 21
5 0 1
9 3 16
1 8 4
4 1
0 3
6 9
2 1
8 7
0 4

样例输出：
-11
-9
-45
-15
17
7

AC代码

#include <iostream>
#include <cstring>
#include <cstdio>

#define MAX 900
#define INF 0x3f3f3f
using namespace std;
struct Edge
{
    int u,v,cost;
}e[MAX];
int dist[MAX];  //最短路径
int m,n;    //边数和顶点数

bool Bellman_Ford(int v0)
{
    int u=v0;
    for(int i=1;i<=n;i++)
        dist[i]=INF;
    dist[u]=0;
    for(int i=1;i<=n;i++)
        for(int j=0;j<m;j++)
            if(dist[e[j].v]>dist[e[j].u]+e[j].cost)
            {
                dist[e[j].v]=dist[e[j].u]+e[j].cost;
            }
    for(int i=0;i<m;i++)
        if(dist[e[i].v]>dist[e[i].u]+e[i].cost)
            return 0;
    return 1;
}

int main()
{    int T;
    cin >> T; 
    while(T--){
        cin>>n>>m;
        for(int i=0;i<m;i++){
            int a,b,c;
            cin >> a >> b >> c;
            e[i].u = a+1;
            e[i].v = b+1;
            e[i].cost = c;
        }
        for(int i = 0 ; i < 6; i++){
            int t1, t2;
            cin >> t1 >> t2;
            t1++;
            t2++;
            Bellman_Ford(t2);
            int res = -dist[t1];
            cout << res << endl;
            e[m].u = t1;
            e[m].v = t2;
            e[m].cost = res;
            m++;
        }
       
    }
    

    return 0;
}