Day4-B-最短路径问题 HDU3790

给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。

Input输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点。n和m为0时输入结束。 
(1<n<=1000, 0<m<100000, s != t)Output输出 一行有两个数， 最短距离及其花费。Sample Input

3 2
1 2 5 6
2 3 4 5
1 3
0 0

Sample Output

9 11

思路：最短路问题，Djikstra即可，注意读入可能有重复边，需要判断一下，代码如下：

const int maxm = 1010;
const int INF = 0x7fffffff;

int N, M, Start, End, G[maxm][maxm], len[maxm][maxm], vis[maxm], cost[maxm], steps[maxm];

struct Node {
    int u, sum, step;
    Node(int _u, int _sum, int _step) : u(_u), sum(_sum), step(_step){}

    bool operator<(const Node &a) const {
        return a.sum < sum || (a.sum == sum && a.step < step);
    }
};

void init() {
    for (int i = 1; i <= N; ++i) {
        cost[i] = steps[i] = INF;
        for (int j = 1; j <= N; ++j)
            G[i][j] = G[j][i] = len[i][j] = len[j][i] = INF;
    }
    memset(vis, 0, sizeof(vis));
}

int main() {
    while(scanf("%d%d",&N,&M) && N + M) {
        init();
        for (int i = 0; i < M; ++i) {
            int t1, t2, t3, t4;
            scanf("%d%d%d%d", &t1, &t2, &t3, &t4);
            if(G[t1][t2] > t3 || (G[t1][t2] == t3 && len[t1][t2] > t4)) {
                G[t1][t2] = G[t2][t1] = t3;
                len[t1][t2] = len[t2][t1] = t4;
            }
        }
        scanf("%d%d", &Start, &End);
        priority_queue<Node> q;
        q.push(Node(Start, 0, 0));
        cost[Start] = steps[Start] = 0;
        while(!q.empty()) {
            Node p = q.top();
            q.pop();
            if(vis[p.u]++)
                continue;
            for (int i = 1; i <= N; ++i) {
                if(G[p.u][i] != INF) {
                    if(cost[i] > cost[p.u] + G[p.u][i] || (cost[i] == cost[p.u] + G[p.u][i]
                        && steps[i] > steps[p.u] + len[p.u][i])) {
                        cost[i] = cost[p.u] + G[p.u][i];
                        steps[i] = steps[p.u] + len[p.u][i];
                        q.push(Node(i, cost[i], steps[i]));
                    }
                }
            }
        }
        printf("%d %d
", cost[End], steps[End]);
    }
    return 0;
}