(Jizhong) 2405. The risk coefficient (danger)] [Floyed

(File IO): input: danger.in output: danger.out
time limit: 1000 ms space constraints: 262144 KB specific restrictions
Goto ProblemSet

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Title Description
$FJ$ in the same boat, the sea has $N(1<=N<=100)$ islands, numbered $1..N$ , and now his task is to order in accordance with a given access $A_1,A_2,….A_M$To explore this $M(2<=M<=10,000)$ islands, has been known risk factor between any two islands, let you explore to find a sequence, just to satisfy your quest sequence contains a given $A_1..A_M$This sequence can be (not necessarily consecutive) so as to minimize the total risk factor.

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Enter the
first $1$ line: two numbers, $N$ and $M$
first $2..M+1$ : Line $i+1$ row represents a given sequence of $i$ islands $A_i$
The first $M+2..N+M+1$ line: each row $N$ integers representing risk factor between islands, must be left diagonal $0$。

Output
minimum risk coefficient output to meet the requirements of

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Sample input
. 3. 4
. 1
2
. 1
. 3
0. 5. 1
. 5 0 2
. 1 0 2

Sample output
7

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Data range limit

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Tips
we can follow $1，3，2，3，1，3$ order to explore, to meet the predetermined sequence is a sequence of words of the sequence, the risk factor is $(1,3)+(3,2)+(2,3)+(3,1)+(1,3)=7$。

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Problem-solving ideas
shortest, $n<=100$， $Floyed$ go from. Very simple. . .

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Code

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
int n,m,a[10010],f[110][110],ans;
int main(){
    freopen("danger.in","r",stdin);
    freopen("danger.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%d",&f[i][j]);
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(i!=j&&i!=k&&j!=k&&f[i][k]+f[k][j]<f[i][j])
                    f[i][j]=f[i][k]+f[k][j];
    for(int i=2;i<=m;i++)
        ans+=f[a[i-1]][a[i]];
    printf("%d",ans);
}