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题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1224
两重循环
判断一下是否可走
再更新一下dp
用pre数组记录路径
最后是用递归输出
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
int cas=1;
int mp[105];
int dp[105],pre[105],load[105][105],i,x,y,j;
void putans(int num)
{
if(num==-1)
return;
putans(pre[num]);
cout<<num<<"->";
}
int main()
{
// std::ios::sync_with_stdio(false);
int t;
cin>>t;
while(t--)
{
memset(load,0,sizeof(load));
memset(mp,0,sizeof(mp));
memset(dp,0,sizeof(dp));
memset(pre,-1,sizeof(pre));
int n,m;
cin>>n;
for(i=1;i<=n;i++)
{
cin>>mp[i];
}
n++;
mp[n]=0;
cin>>m;
for(i=1;i<=m;i++)
{
cin>>x>>y;
load[x][y]=1;
}
for(j=1;j<=n;j++)
{
for(i=1;i<=n;i++)
{
if(load[i][j]&&dp[j]<dp[i]+mp[i])
{
dp[j]=dp[i]+mp[i];
pre[j]=i;
}
}
}
if(cas>1)cout<<endl;
cout<<"CASE "<<cas++<<"#"<<endl;
cout<<"points : "<<dp[n]<<endl;
cout<<"circuit : 1->";
putans(pre[n]);
cout<<"1"<<endl;
}
return 0;
}