bellman-ford(可判负权回路＋记录路径)

Posted 2020-09-16 远搏

 

 

#include<iostream>
#include<cstdio>
using namespace std;

#define MAX 0x3f3f3f3f
#define N 1010
int nodenum, edgenum, original; //点，边，起点

typedef struct Edge //边
{
    int u, v;
    int cost;
}Edge;

Edge edge[N];
int dis[N], pre[N];

bool Bellman_Ford()
{
    for(int i = 1; i <= nodenum; ++i) //初始化
        dis[i] = (i == original ? 0 : MAX);
    for(int i = 1; i <= nodenum - 1; ++i)
        for(int j = 1; j <= edgenum; ++j)
            if(dis[edge[j].v] > dis[edge[j].u] + edge[j].cost) //松弛（顺序一定不能反~）
            {
                dis[edge[j].v] = dis[edge[j].u] + edge[j].cost;
                pre[edge[j].v] = edge[j].u;
            }
            bool flag = 1; //判断是否含有负权回路
            for(int i = 1; i <= edgenum; ++i)
                if(dis[edge[i].v] > dis[edge[i].u] + edge[i].cost)
                {
                    flag = 0;
                    break;
                }
                return flag;
}

void print_path(int root) //打印最短路的路径（反向）
{
    while(root != pre[root]) //前驱
    {
        printf("%d-->", root);
        root = pre[root];
    }
    if(root == pre[root])
        printf("%d\n", root);
}

int main()
{
    scanf("%d%d%d", &nodenum, &edgenum, &original);
    pre[original] = original;
    for(int i = 1; i <= edgenum; ++i)
    {
        scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].cost);
    }
    if(Bellman_Ford())
        for(int i = 1; i <= nodenum; ++i) //每个点最短路
        {
            printf("%d\n", dis[i]);
            printf("Path:");
            print_path(i);
        }
    else
        printf("have negative circle\n");
    return 0;
}

4 6 1
1 2 20
1 3 5
4 1 -200
2 4 4
4 2 4
3 4 2

和：

4 6 1
1 2 2
1 3 5
4 1 10
2 4 4
4 2 4
3 4 2