【最短路-spfa算法】Road Construction Aizu - 2249

比眉伴天荒 2022-05-31 01:42 181阅读 0赞

Think:  
1知识点：最短路-spfa算法  
2题意：n个城市m条路，每条路的信息包括起点(u)、终点(v)、长度(d)、花费(c)，询问在保证1号城市到其它城市距离最短的条件下，连通n个城市的最少花费。  
2.1数据范围：  
1 <= n <= 1e4  
0 <= m <= 2e4  
1 <= d <= 1000  
1 <= c <= 1000  
3错误反思：  
3.1：spfa算法求最短路中，手动模拟队列时注意如果不是用循环队列需要注意队列的大小(因为虽然在某一时刻队列中不会出现重复的元素，但某一元素可能会多次入队更新最短路)

[vjudge题目链接][vjudge]

以下为Accepted代码——循环队列

#include <cstdio>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    
    const int inf = 0x3f3f3f3f;
    const int N = 10414;
    const int M = 40414;
    
    struct Edge{
        int u, v, d, c;
        int next;
    }edge[M];
    
    int cnt, head[N];
    int vis[N], dis[N], cost[N];
    int op, tp, link[N];
    
    void add_edge(int u, int v, int d, int c);
    void spfa(int n);
    
    int main(){
        int n, m, i, u, v, d, c;
        while(~scanf("%d %d", &n, &m) && (n || m)){
            cnt = 0;
            memset(head, -1, sizeof(head));
            for(i = 0; i < m; i++){
                scanf("%d %d %d %d", &u, &v, &d, &c);
                add_edge(u, v, d, c);
                add_edge(v, u, d, c);
            }
            spfa(n);
        }
        return 0;
    }
    void add_edge(int u, int v, int d, int c){
        edge[cnt].u = u;
        edge[cnt].v = v;
        edge[cnt].d = d;
        edge[cnt].c = c;
        edge[cnt].next = head[u];
        head[u] = cnt++;
        return;
    }
    void spfa(int n){
        int i, u, v, d, c, sum;
        memset(vis, 0, sizeof(vis));
        memset(dis, inf, sizeof(dis));
        memset(cost, inf, sizeof(cost));
        op = tp = 0;
        vis[1] = 1, dis[1] = 0, cost[1] = 0;
        link[tp++] = 1;
        while(true){
            op %= N;
            tp %= N;
            if(op == tp) break;
            u = link[op++];
            vis[u] = 0;
            for(i = head[u]; ~i; i = edge[i].next){
                v = edge[i].v, d = edge[i].d, c = edge[i].c;
    
                if(dis[v] > dis[u] + d || (dis[v] == dis[u] + d && cost[v] > c)){
                    dis[v] = dis[u] + d;
                    cost[v] = c;
                    if(!vis[v]){
                        vis[v] = 1;
                        link[tp++] = v;
                    }
                }
            }
        }
        sum = 0;
        for(i = 1; i <= n; i++){
            sum += cost[i];
        }
        printf("%d\n", sum);
        return;
    }

[vjudge]: https://vjudge.net/problem/Aizu-2249