1030. Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权)
来源:互联网 发布:c2c 源码 编辑:程序博客网 时间:2024/05/29 10:08
1030. Travel Plan (30)
A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.
Input Specification:
Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (<=500) is the number of cities (and hence the cities are numbered from 0 to N-1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:
City1 City2 Distance Cost
where the numbers are all integers no more than 500, and are separated by a space.
Output Specification:
For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.
Sample Input
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20
Sample Output
0 2 3 3 40
题目大意:求起点到终点的最短路径最短距离和花费,要求首先路径最短,其次花费最少,要输出完整路径。
分析:Dijksta + DFS。 Dijkstra记录路径pre数组,然后用dfs求最短的一条mincost以及它的路径path,最后输出path数组和mincost
注意路径path因为是从末端一直压入push_back到path里面的,所以要输出路径的时候倒着输出
#include <cstdio>#include <algorithm>#include <vector>using namespace std;int n, m, s, d;int e[510][510], dis[510], cost[510][510];vector<int> pre[510];bool visit[510];const int inf = 99999999;vector<int> path, temppath;int mincost = inf;void dfs(int v) { if(v == s) { temppath.push_back(v); int tempcost = 0; for(int i = temppath.size() - 1; i > 0; i--) { int id = temppath[i], nextid = temppath[i-1]; tempcost += cost[id][nextid]; } if(tempcost < mincost) { mincost = tempcost; path = temppath; } temppath.pop_back(); return ; } temppath.push_back(v); for(int i = 0; i < pre[v].size(); i++) dfs(pre[v][i]); temppath.pop_back();}int main() { fill(e[0], e[0] + 510 * 510, inf); fill(dis, dis + 510, inf); scanf("%d%d%d%d", &n, &m, &s, &d); for(int i = 0; i < m; i++) { int a, b; scanf("%d%d", &a, &b); scanf("%d", &e[a][b]); e[b][a] = e[a][b]; scanf("%d", &cost[a][b]); cost[b][a] = cost[a][b]; } pre[s].push_back(s); dis[s] = 0; for(int i = 0; i < n; i++) { int u = -1, minn = inf; for(int j = 0; j < n; j++) { if(visit[j] == false && dis[j] < minn) { u = j; minn = j; } } if(u == -1) break; visit[u] = true; for(int v = 0; v < n; v++) { if(visit[v] == false && e[u][v] != inf) { if(dis[v] > dis[u] + e[u][v]) { dis[v] = dis[u] + e[u][v]; pre[v].clear(); pre[v].push_back(u); } else if(dis[v] == dis[u] + e[u][v]) { pre[v].push_back(u); } } } } dfs(d); for(int i = path.size() - 1; i >= 0; i--) printf("%d ", path[i]); printf("%d %d", dis[d], mincost); return 0;}
- 1030. Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权)
- 1030. Travel Plan (30)-PAT甲级真题
- PAT甲级1030. Travel Plan(30)
- 【PAT甲级】1030. Travel Plan (30)
- PAT甲级练习1030. Travel Plan (30)
- PAT甲级1030. Travel Plan (30)
- 1030. Travel Plan (30) PAT 甲级
- PAT 甲级 1030. Travel Plan (30)
- PAT甲级 1030. Travel Plan (30)
- PAT 1030. Travel Plan (30)(Dijkstra,最短路径的同时计算最小奥cost)
- PAT (Advanced Level) 1030. Travel Plan (30) Dijkstra最短路径
- 1018. Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS)
- 1111. Online Map (30)-PAT甲级真题(Dijkstra + DFS)
- 1030. Travel Plan (30) - Dijkstra
- 1072. Gas Station (30)-PAT甲级真题(Dijkstra)
- 【PAT 1030】Travel Plan 最短路径Dijkstra
- PAT 1030 Travel Plan(单源最短路径+优化Dijkstra)
- 1030. Travel Plan (30)-PAT
- MySQL自定义函数
- C/C++之主函数获取子函数变量地址
- TOJ 1179.Game of Connections(大数模板)
- java用substring函数截取string中一段字符串
- 快速排序quicksort
- 1030. Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权)
- Could not write content: Infinite recursion (StackOverflowError)
- c#中decimal的去0显示
- LOG4NET写日志
- 高并发量网站解决方案
- CSU - 1212 中位数
- Core Animation - 发光的太阳(附高校设置图片圆角和变圆的方法)
- Android Studio如何集成Genymotion
- CPU卡发卡步骤