NOJ——[1176] Exchange Rate

最新推荐文章于 2024-10-14 15:27:31 发布

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最新推荐文章于 2024-10-14 15:27:31 发布

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本文介绍了一个关于货币兑换的问题，探讨了如何找到从一种货币兑换回自身时的最小损耗路径。通过使用动态规划的方法，实现了一个算法来解决这个问题。

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问题描述
You know, every country's currency has an exchange rate to another country.
For example, 'Dollir' : 'RNB' is 1 : 6.
And exchanging from one currency to another currency, it will have a wastage.
Now give some currency and their each wastage from one currency exchanges to another. For every currency, you should tell me when it exchanges to other currency and finally exchanges to itself, the minimum wastage it will cost.
输入
This problem contains several cases.
The first line of each case is an integer N (2 <= N <= 100).
Then follows a decimal matrix(N * N). The Cell(i, j) indicates the wastage exchanging from ith currency to jth currency. (1.00 < Cell(i, j) <= 100.00).
You may consider that every currency can't exchange to itself directly, so every Cell(i, i) will be -1.
输出
For each case, you may output N lines.
Each line is an decimal, indicates exchanging from ith currency and finally it exchanges to itself, the minimum wastage it will cost.
Two decimal places.
样例输入

4
-1 1.00 2.00 10.00
1.00 -1 2.00 30.00
1.00 1.00 -1 50.00
50.00 50.00 1.0 -1

样例输出
```
2.00
2.00
3.00
12.00
```
提示
```
无
```
来源
```
XadillaX
```

求各个点回到自身的最短路。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f

using namespace std;

double dp[105][105];

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)
		  for(int j=1;j<=n;j++)
		  {
  			scanf("%lf",&dp[i][j]);
  			if(i==j)
  			  dp[i][j]=inf;
  		  }
        for(int k=1;k<=n;k++)
           for(int i=1;i<=n;i++)
             for(int j=1;j<=n;j++)
               dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
        for(int i=1;i<=n;i++)
          printf("%.2f\n",dp[i][i]);
	}
	return 0;
}