F - Wormholes——最短路_spfa()算法+前向星

约定不等于承诺〃 2022-06-12 06:10 201阅读 0赞

Think:  
1知识点：最短路\_spfa()算法+前向星  
2反思：全局变量不要重复定义，memset()初始化参数不要传错

[建议参考题目分析][Link 1]

[vjudge题目链接][vjudge]

以下为Accepted代码

#include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    
    using namespace std;
    
    const int inf = 0x3f3f3f3f;
    const int N = 5e2 + 4;
    const int M = 3e3 + 4;
    
    struct node {
        int v;
        int w;
        int next;
    }edge[M*2];
    
    int n, cnt, head[N], dis[N], vis[N], num[N];
    
    void add_edge(int u, int v, int w);
    bool spfa(int x);
    
    int main(){
        int T, m, wor, i, u, v, w;
        scanf("%d", &T);
        while(T--){
            cnt = 0;
            memset(head, -1, sizeof(head));
            scanf("%d %d %d", &n, &m, &wor);
            for(i = 1; i <= m; i++){
                scanf("%d %d %d", &u, &v, &w);
                add_edge(u, v, w);
                add_edge(v, u, w);
            }
            for(i = 1; i <= wor; i++){
                scanf("%d %d %d", &u, &v, &w);
                add_edge(u, v, -w);
            }
            bool flag = false;
            for(i = 1; i <= n; i++){
                if(spfa(i)){
                    flag = true;
                    break;
                }
            }
            if(flag)
                printf("YES\n");
            else
                printf("NO\n");
        }
        return 0;
    }
    void add_edge(int u, int v, int w){
        edge[cnt].v = v;
        edge[cnt].w = w;
        edge[cnt].next = head[u];
        head[u] = cnt++;
    }
    bool spfa(int x){
        queue <int> q;
        while(!q.empty()){
            q.pop();
        }
        memset(dis, inf, sizeof(dis));
        memset(vis, 0, sizeof(vis));
        memset(num, 0, sizeof(num));
        dis[x] = 0, vis[x] = 1;
        q.push(x);
        num[x]++;
        while(!q.empty()){
            int u = q.front();
            q.pop();
            vis[u] = 0;
            for(int i = head[u]; ~i; i = edge[i].next){
                int v = edge[i].v;
                if(dis[u] + edge[i].w < dis[v]){
                    dis[v] = dis[u] + edge[i].w;
                    if(!vis[v]){
                        vis[v] = 1;
                        q.push(v);/*松弛点入队*/
                        num[v]++;
                    }
                    if(num[v] >= n){
                        return true;
                    }
                }
            }
        }
        return false;
    }

[Link 1]: http://www.cnblogs.com/liuxin13/p/4653983.html
[vjudge]: https://vjudge.net/contest/163006#problem/F