九度oj题目1008：最短路径问题

题目描述：

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。

输入：                       

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t。n和m为0时输入结束。 (1<n<=1000, 0<m<100000, s != t)

输出：                       

输出 一行有两个数， 最短距离及其花费。

样例输入：                       

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

样例输出：                       

9 11
通过这道题复习了一下迪杰特斯拉算法，算法的精髓在于每回找一个距离集合最近的点加到集合中，直到不能加入更多的点为止
代码如下：

 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #define MAX 999999999
 #define N 1005
 #define M 100005

 struct Way {
     int distance;
     int cost;
 };

 Way map[N][N];//邻接矩阵
 Way dis[N];//起点到各个点的距离表
 int flag[N];//判别某点是否在已包括点的集合中

 void dijkstra(int start, int end,int n) {
     memset(flag,,sizeof(int)*N);
     //初始化距离
     for(int i = ; i <= n; i++) {
         dis[i] = map[start][i];
     }
     dis[start].distance = ;//该点到自己的距离为0
     dis[start].cost = ;//到自己的花费为0
     flag[start] = ;//把自己包括到可以到达的点中

     for(int i = ; i < n; i++) {
         //总共需要加入n-1个点，所以循环n-1次
         int index = ;
         int min = MAX;
         for(int j = ; j <= n; j++) {
             if(flag[j] ==  && dis[j].distance < min) {
                 index = j;
                 min = dis[j].distance;
             }
         }
         //find the min index
         flag[index] = ;

         for(int j = ; j <= n; j++) {
             if(flag[j] ==  && (dis[index].distance + map[index][j].distance < dis[j].distance)) {
                 dis[j].distance = dis[index].distance + map[index][j].distance;
                 dis[j].cost = dis[index].cost + map[index][j].cost;
             }
             else if(flag[j] ==  && (dis[index].distance + map[index][j].distance == dis[j].distance)){
                 if(dis[j].cost > dis[index].cost + map[index][j].cost) {
                     dis[j].cost = dis[index].cost + map[index][j].cost;
                 }
             }
         }
     }

 }

 int main(int argc, char const *argv[])
 {
     int n,m;
     int source, des;
     scanf("%d %d",&n,&m);
     while(n !=  && m != ) {
         for(int i = ; i <= n; i++) {
             for(int j = ; j <= n; j++) {
                 map[i][j].distance = MAX;
                 map[i][j].cost = MAX;
             }
         }
         int pointA,pointB,disc,cost;
         for(int i = ; i < m; i++) {
             scanf("%d %d %d %d",&pointA,&pointB,&disc,&cost);
             map[pointA][pointB].distance = disc;
             map[pointB][pointA].distance = disc;
             map[pointA][pointB].cost = cost;
             map[pointB][pointA].cost = cost;
         }
         scanf("%d %d",&source,&des);

         dijkstra(source,des,n);
         printf("%d %d\n",dis[des].distance,dis[des].cost);
         scanf("%d %d",&n,&m);
     }
     return ;
 }

最近尝试了一下用优先队列来解决问题，感觉写得很别扭，不过还是写出来了，速度提高不少，代码如下

 #include <cstdio>
 #include <algorithm>
 #include <vector>
 #include <queue>

 #define MAX_V 1002
 #define inf 999999999
 using namespace std;

 struct Edge
 {
 public:
     int to;
     int dis;
     int cost;
     Edge(int to, int dis,int cost) {
         this->to = to;
         this->dis = dis;
         this->cost = cost;
     }
 };

 struct cmp{
     bool operator()(const Edge &a,const Edge &b){
         if(a.dis == b.dis) {
             return a.cost - b.cost;
         }
         else {
             return a.dis - b.dis;
         }
     }
 };
 typedef pair<int , int> P;//first dis, sencond cost
 P dist[MAX_V];
 vector<Edge> G[MAX_V];
 priority_queue<Edge, vector<Edge>, cmp> que;

 int main(int argc, char const *argv[])
 {
     int n,m;
     int source, des;
     freopen("input.txt","r",stdin);
     while(scanf("%d %d",&n,&m) != EOF && (n !=  && m != )) {
         for(int i = ; i <= n; i++) {
             G[i].clear();
         }
         for(int i = ; i < m; i++) {
             int a, b, disc, cost;
             scanf("%d %d %d %d",&a,&b,&disc,&cost);
             G[a].push_back(Edge(b,disc,cost));
             G[b].push_back(Edge(a,disc,cost));
         }
         scanf("%d %d",&source,&des);
         fill(dist,dist+n+,P(inf,inf));
         dist[source] = P(,);
         que.push(Edge(source,,));

         while(!que.empty()) {
             Edge t = que.top(); que.pop();
             int v = t.to, d = t.dis, c = t.cost;
             if(dist[v].first < d) {
                 continue;
             }
             for(int i = ; i < G[v].size(); i++) {
                 Edge &e = G[v][i];
                 int d2 = d + e.dis;
                 int c2 = c + e.cost;

                 if(dist[e.to].first > d2 || (dist[e.to].first == d2 && dist[e.to].second > c2)) {
                     dist[e.to] = P(d2,c2);
                     que.push(Edge(e.to,d2,c2));
                 }
             }
         }
         printf("%d %d\n",dist[des].first,dist[des].second);

     }
     return ;
 }