1111. Online Map (30)[dijkstra算法]

1. 原题： https://www.patest.cn/contests/pat-a-practise/1111

2. 思路：

题意：
给定图含有时间及距离。
求单源最短路径。
要求：
1. 输出距离最短路径。存在多条时，选择时间最短的。
2. 输出时间最短的路径。存在多条时，选择结点数最少的。


思路：
题意不难。显然dijkstra算法。
对距离和时间分别dijkstra就可以了。
我写了两个dijkstra函数（代码比较长，好在第二个复制改了下）。
也可以写一个，然后调用不同参数，这样代码少。都可以。
已AC。

3. 源码：

#include<iostream>#include<vector>using namespace std;const int Max = 500;//最大结点数const int INF = 0x7FFFFFFF;//无穷大int N, M, s, d;//分别为结点数， 路径数， 起点，目的地。int len[Max][Max], dtime[Max][Max];//分别为路径长度的图和时间花费的图int collected[Max] = { 0 };//记录是否已经遍历int path[Max];//记录路径int dist[Max];//记录起点到某点的花费int findMin();//找出未收录的最小长度void dijkstra_len();//对路径长度图进行dijkstravoid dijkstra_time();//对时间花费图进行dijkstravoid generatePath(vector<int> &v, int index);//生成最短路径void print(vector<int> &v);//输出路径int main(void){//freopen("in.txt", "r", stdin);scanf("%d %d", &N, &M);for (int i = 0; i < N; i++)//图进行初始化{for (int j = 0; j < N; j++)len[i][j] = dtime[i][j] = INF;}for (int i = 0; i < M; i++)//读入数据{int start, end, flag, t_len, t_time;scanf("%d %d %d %d %d", &start, &end, &flag, &t_len, &t_time);if (flag == 0)//说明是双向{len[start][end] = len[end][start] = t_len;dtime[start][end] = dtime[end][start] = t_time;}else//单向路径{len[start][end] = t_len;dtime[start][end] = t_time;}}scanf("%d %d", &s, &d);//起点， 终点dijkstra_len();//对路径长度图进行dijkstravector<int> timepath, lenpath;//分别存储两个输出的路径int lenWgt = dist[d];//距离总开销generatePath(lenpath, d);//生成路径dijkstra_time();int timeWgt = dist[d];generatePath(timepath, d);if (timepath == lenpath)//相同路径，输出一个即可{printf("Distance = %d; Time = %d: ", lenWgt, timeWgt);print(timepath);}else{printf("Distance = %d: ", lenWgt);print(lenpath);printf("Time = %d: ", timeWgt);print(timepath);}return 0;}void dijkstra_len()//对路径长度图进行dijkstra{int total[Max];//记录到达某点时的总时间for (int i = 0; i < N; i++)//初始化{total[i] = dtime[s][i];dist[i] = len[s][i];path[i] = -1;//这里的值是存储父结点collected[i] = 0;}dist[s] = 0;total[s] = 0;collected[s] = 1;while (1)//经典算法{int v = findMin();if (v == -1)break;collected[v] = 1;for (int w = 0; w < N; w++){if (collected[w] == 0 && len[v][w] < INF){if ((dist[v] + len[v][w]) < dist[w]){dist[w] = dist[v] + len[v][w];path[w] = v;total[w] = total[v] + dtime[v][w];}else if ((dist[v] + len[v][w]) == dist[w]){if ((total[v] + dtime[v][w]) < total[w]){path[w] = v;total[w] = total[v] + dtime[v][w];}}}}}return;}void dijkstra_time()//对时间花费图进行dijkstra{int cnt[Max] = { 0 };//记录到达某点的总结点数for (int i = 0; i < N; i++){dist[i] = dtime[s][i];path[i] = -1;collected[i] = 0;}dist[s] = 0;collected[s] = 1;while (1){int v = findMin();if (v == -1)break;collected[v] = 1;for (int w = 0; w < N; w++){if (collected[w] == 0 && dtime[v][w] < INF){if ((dist[v] + dtime[v][w]) < dist[w]){dist[w] = dist[v] + dtime[v][w];path[w] = v;cnt[w] = cnt[v] + 1;}else if ((dist[v] + dtime[v][w]) == dist[w]){if ((cnt[v] + 1) < cnt[w]){path[w] = v;cnt[w] = cnt[v] + 1;}}}}}return;}int findMin()//找出未收录的最小长度{int min_val = INF;int min_id = -1;for (int i = 0; i < N; i++){if (collected[i] == 0 && dist[i] < min_val){min_val = dist[i];min_id = i;}}return min_id;//-1表dijkstra算法结束}void generatePath(vector<int> &v, int index)//递归把结点压入容器{if (index == -1){v.push_back(s);return;}generatePath(v, path[index]);v.push_back(index);return;}void print(vector<int> &v)//输出路径{for (int i = 0; i < v.size(); i++){if (i != 0)printf(" -> ");printf("%d", v[i]);}printf("\n");}