【最短路-spfa算法】Road Construction Aizu

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

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

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;
}