bzoj1624 [Usaco2008 Open] Clear And Present Danger 寻宝之路

2018年01月13日 算法 ⁄ 共 1536字 ⁄ 字号 评论关闭

Description

农夫约翰正驾驶一条小艇在牛勒比海上航行．
海上有N(1≤N≤100)个岛屿，用1到N编号．约翰从1号小岛出发，最后到达N号小岛．一

Input

第1行输入N和M，之后M行一行一个整数表示A序列，之后输入一个NxN的方阵，表示两两岛屿之间航线的危险指数．数据保证Dij=Dji，Dii=0．

最小的危险指数和．

Sample Input

3 4

1

2

1

3

0 5 1

5 0 2

1 2 0

INPUT DETAILS:

There are 3 islands and the treasure map requires Farmer John to

visit a sequence of 4 islands in order: island 1, island 2, island

1 again, and finally island 3. The danger ratings of the paths are

given: the paths (1, 2); (2, 3); (3, 1) and the reverse paths have

danger ratings of 5, 2, and 1, respectively.

Sample Output

7

OUTPUT DETAILS:

He can get the treasure with a total danger of 7 by traveling in

the sequence of islands 1, 3, 2, 3, 1, and 3. The cow map's requirement

(1, 2, 1, and 3) is satisfied by this route. We avoid the path

between islands 1 and 2 because it has a large danger rating.

```#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x*=10;x+=ch-'0';ch=getchar();}
return x*f;
}
int dist[101][101];
int a[10001];
int ans;
int main()
{
for (int i=1;i<=m;i++)
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)