bzoj1624 [Usaco2008 Open] Clear And Present Danger 寻宝之路
Description
Input
Output
Sample Input
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
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.
用Floyd预处理出两两之间的最小边权和,然后按题目给的顺序走一遍
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;
inline int read()
{
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()
{
int n=read(),m=read();
for (int i=1;i<=m;i++)
a[i]=read();
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
dist[i][j]=read();
for (int k=1;k<=n;k++)
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);
for (int i=2;i<=m;i++)ans+=dist[a[i-1]][a[i]];
cout<<ans;
}
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