Hdu-2680 Choose the best route

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2680

题目大意：

给你一个有向图，一个起点集合，一个终点，求最短路。。。。

解题思路：

1.自己多加一个超级源点，把起点集合连接到超级源点上，然后将起点与超级源点的集合的路径长度设为0，这样就称为一个n+1个点的单源最短路算法。。。。。

2.反向图+终点的Dijkstra，然后记录最小值。

思路1的代码如下：

 #include
 #include
 #include
 using namespace std;

 #define MAX 0x3f3f3f3f
 #define N 1010

 int num, road, destination;
 int map[N][N], dis[N];
 bool visit[N];

 void Dijkstra()
 {
     int temp, k;
     memset(visit, 0, sizeof(visit));
     for(int i = 0; i <= num; ++i) //n+1个点的Dijkstra
         dis[i] = map[0][i];
     dis[0] = 0;
     visit[0] = 1;
     for(int i = 0; i <= num; ++i)
     {
         temp = MAX;
         for(int j = 0; j <= num; ++j)
             if(!visit[j] && temp > dis[j])
                 temp = dis[k = j];
         if(temp == MAX) break;
         visit[k] = 1;
         for(int j = 0; j <= num; ++j)
             if(!visit[j] && dis[j] > dis[k] + map[k][j])
                 dis[j] = dis[k] + map[k][j];
     }
 }

 int main()
 {
     int x, y, cost, exp, cost1;
     while(scanf("%d%d%d", &num, &road, &destination) != EOF)
     {
         memset(map, MAX, sizeof(map));
         for(int i = 1; i <= road; ++i)
         {
             scanf("%d%d%d", &x, &y, &cost);
             if(cost < map[x][y])
             map[x][y] = cost;
         }
         scanf("%d", &exp);
         for(int i = 1; i <= exp; ++i) //巧妙之处，加超级源点
         {
             scanf("%d", &cost1);
             map[0][cost1] = 0;
         }
         Dijkstra();
         if(dis[destination] == MAX) printf("-1\n");
         else printf("%d\n", dis[destination]);
     }
     return 0;
 }

思路2代码如下：

 #include
 #include
 #include
 using namespace std;

 #define MAX 0x3f3f3f3f
 #define N 1010

 int num, road, destination;
 int map[N][N], dis[N], oppo[N];
 bool visit[N];

 void Dijkstra(int start)
 {
     int temp, k;
     memset(visit, 0, sizeof(visit));
     for(int i = 1; i <= num; ++i)
         dis[i] = map[start][i];
     dis[start] = 0;
     visit[start] = 1;
     for(int i = 1; i <= num; ++i)
     {
         temp = MAX;
         for(int j = 1; j <= num; ++j)
             if(!visit[j] && temp > dis[j])
                 temp = dis[k = j];
         if(temp == MAX) break;
         visit[k] = 1;
         for(int j = 1; j <= num; ++j)
             if(!visit[j] && dis[j] > dis[k] + map[k][j])
                 dis[j] = dis[k] + map[k][j];
     }
 }

 int main()
 {
     int x, y, cost, exp, minn;
     while(scanf("%d%d%d", &num, &road, &destination) != EOF)
     {
         memset(map, MAX, sizeof(map));
         for(int i = 1; i <= road; ++i)
         {
             scanf("%d%d%d", &x, &y, &cost);
             if(cost < map[y][x]) //巧妙之处：反向图
             map[y][x] = cost;
         }
         scanf("%d", &exp);
         for(int i = 1; i <= exp; ++i)
             scanf("%d", &oppo[i]);
         Dijkstra(destination);
         minn = MAX;
         for(int i = 1; i <= exp; ++i) //起点找到终点最小值
             if(dis[oppo[i]] < minn)
                 minn = dis[oppo[i]];
         if(minn == MAX) printf("-1\n");
         else printf("%d\n", minn);
     }
     return 0;
 }