[九度OJ]最短路径

来源：互联网发布：与sqlserver建立连接编辑：程序博客网时间：2024/04/30 05:20

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：5046

解决：766

题目描述：: N个城市，标号从0到N-1，M条道路，第K条道路（K从0开始）的长度为2^K，求编号为0的城市到其他城市的最短距离

输入：: 第一行两个正整数N（2<=N<=100）M(M<=500),表示有N个城市，M条道路
接下来M行两个整数，表示相连的两个城市的编号

输出：: N-1行，表示0号城市到其他城市的最短路，如果无法到达，输出-1，数值太大的以MOD 100000 的结果输出。

样例输入：

4 41 22 31 30 1

样例输出：

来源：
2010年上海交通大学计算机研究生机试真题

Dijkstra: 还有问题
import java.math.*;import java.util.*;public class Main { static final BigInteger TWO = new BigInteger("2"); static final BigInteger MAX = Main.TWO.pow(501); public static void Dijkstra(BigInteger[][] dist, BigInteger[] cost, int n) { int[] tag = new int[n]; tag[0] = 1; int cnt = 1; while (cnt != n) { BigInteger minCost = Main.MAX; int pos = 0; for (int i = 0; i < n; i++) { if (tag[i]==0 && cost[i].compareTo(minCost)<0) { minCost = cost[i]; pos = i; } } if (pos == 0) return; tag[pos] = 1; cnt++; cost[pos] = minCost; for (int i = 0; i < n; i++) { if (tag[i] == 0) { BigInteger tmpCost = (dist[pos][i]==MAX ? MAX:cost[pos].add(dist[pos][i])); if (tmpCost.compareTo(cost[i]) < 0) { cost[i] = tmpCost; } } } } } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int n, m; n = scanner.nextInt(); m = scanner.nextInt(); BigInteger[][] dist = new BigInteger[n][n]; BigInteger[] cost = new BigInteger[n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (i == j) dist[i][j] = BigInteger.ZERO; else dist[i][j] = Main.MAX; } } int xpos, ypos; BigInteger currVal = new BigInteger("1"); for (int i = 0; i < m; i++) { xpos = scanner.nextInt(); ypos = scanner.nextInt(); dist[xpos][ypos] = currVal; currVal = currVal.multiply(Main.TWO); } cost[0] = BigInteger.ZERO; for (int i = 1; i < n; i++) { cost[i] = dist[0][i]; } Dijkstra(dist, cost, n); for (int i = 1; i < n; i++) { if (cost[i].compareTo(MAX) < 0) System.out.println(cost[i]); else System.out.println("-1"); } scanner.close(); }}

SPFA: 非本题代码
#include <iostream>#include <cstring>#include <string>#include <fstream>#include <vector>#include <queue>#include <cmath>#define MAXN 1005using namespace std;double city[MAXN][2];int vis[MAXN];int path[MAXN];int A = 144;double orig_cost[MAXN];struct Link{int end;double cost;};vector<Link> vLink[MAXN];double dist(int sp, int ep){double x = city[sp][0] - city[ep][0];double y = city[sp][1] - city[ep][1];return sqrt(double(x*x + y*y));}void output_path(int pos){if (path[pos] != 1)output_path(path[pos]);printf("%d->", path[pos]);}double SPFA(int src,int target){queue<int> QL;QL.push(src);while (!QL.empty()){int origp = QL.front();QL.pop();vis[origp] = 0;for (int i = 0; i < (int)vLink[origp].size(); i++){int currp = vLink[origp][i].end;double tmp_cost = orig_cost[origp] + vLink[origp][i].cost;if (tmp_cost < orig_cost[currp]){orig_cost[currp] = tmp_cost;if (vis[currp] == 0){vis[currp] = 1;QL.push(currp);}path[currp] = origp;}}}return orig_cost[target];}int main(int argc,char **argv){/*if (argc != 3){cerr << "argc error!" << endl;exit(1);}*/freopen("Cities(144).txt", "r", stdin);double px, py;int pcnt;while (scanf("%d %lf %lf\n", &pcnt, &px, &py) != EOF){city[pcnt][0] = px;city[pcnt][1] = py;}for (int i = 2; i <= pcnt; i++)orig_cost[i] = 1e10;memset(vis, 0, sizeof(vis));memset(path, 0, sizeof(path));freopen("Cities(144)link.txt", "r", stdin);int lstart, lend;while (scanf("%d %d\n", &lstart, &lend) != EOF){double c = dist(lstart, lend);vLink[lstart].push_back(Link{ lend, c });vLink[lend].push_back(Link{ lstart, c });}double ans=SPFA(1,A);output_path(A);printf("%d\n%lf\n", A, ans);return 0;}

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