1624: [Usaco2008 Open] Clear And Present Danger 尋寶之路
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 763 Solved: 499
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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.
HINT
Source
題解:Floyd就可以過。。用Floyed跑一遍,求出每兩點之間的最短的路徑,最後統計一下答案就好了。。
代碼:
#include<cstdio>
using namespace std;
int f[101][101];
int b[100001];
int n,m,ans;
int read()
{
int f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int main()
{
n=read();m=read();
for(int i=1;i<=m;i++)
b[i]=read();
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
f[i][j]=read();
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(f[i][k]+f[k][j]<f[i][j])f[i][j]=f[i][k]+f[k][j];
for(int i=0;i<=m;i++)
ans=ans+f[b[i]][b[i+1]];
printf("%d",ans);
return 0;
}