遍历 0 ≤ i < n − 1 0 \le i \lt n-1 0≤i<n−1,判断多少个i满足 s t r [ i ] str[i] str[i]和 s t r [ n − 1 ] str[n-1] str[n−1]奇偶性相同
int main(){
string s;cin>>s;
int ans = 0;
for(int i=0;i<s.size()-1;++i)
if((s[i]-'0'+s[s.size()-1]-'0')%2==0)
ans++;
cout<<ans<<endl;
return 0;
}
计算所有字符变为字符c的代价
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\min_c \sum_{i=1}^n min(abs(i - c), 26 - abs(i - c))
cmini=1∑nmin(abs(i−c),26−abs(i−c))
条件转化为:
按照上述规则进行dfs剪枝即可找到答案
#include
using namespace std;
string str;
bool ok;
void dfs(int x)
{
if(ok)return;
if(x==str.size())
{
cout<<str<<endl;
ok = 1;
return;
}
if(str[x]=='?')
{
str[x] = '0';dfs(x);
str[x] = '1';dfs(x);
str[x] = '2';dfs(x);
str[x] = '?';
}
else
{
if(x>0 && str[x]==str[x-1])
return;
if(x>1 && (str[x]-'0'+str[x-1]-'0'+str[x-2]-'0')%2==1)
return;
dfs(x+1);
}
}
int main()
{
cin>>str;
dfs(0);
if(ok==0)
cout<<-1<<endl;
}
经典的树上dp
令
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j的父亲节点,有
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dp[i][0] = \sum max(dp[j][1],dp[j][0]) \\ dp[i][1] = \max_j (\sum max(dp[j][1],dp[j][0]))-max(dp[j][1],dp[j][0])+w_{ij}+dp[j][0]
dp[i][0]=∑max(dp[j][1],dp[j][0])dp[i][1]=jmax(∑max(dp[j][1],dp[j][0]))−max(dp[j][1],dp[j][0])+wij+dp[j][0]
#include
#include
#define ll long long
using namespace std;
vector<pair<int,int>>G[100010];
ll dp[100010][2];
void dfs(int u,int fa)
{
ll mx = 0;
for(auto [v,w]:G[u])
{
if(v==fa)continue;
dfs(v,u);
dp[u][0] += max(dp[v][0],dp[v][1]);
mx = max(dp[v][0] +w-max(dp[v][0],dp[v][1]),mx);
}
dp[u][1] =dp[u][0]+mx;
}
int main()
{
int n;cin>>n;
for(int i=1;i<n;++i)
{
int x,y,w;cin>>x>>y>>w;
G[x].push_back({y,w});
G[y].push_back({x,w});
}
dfs(1,-1);
cout<<max(dp[1][0],dp[1][1])<<endl;
}