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; }