[九度OJ]最短路径

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 4
1 2
2 3
1 3
0 1

样例输出：

8
9
11

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
;
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

;
BigInteger[] cost = new BigInteger
;
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 1005
using 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;
}