B题

https://ac.nowcoder.com/acm/contest/948/D

思路:前向星存边,奇数边就是反向边,dfs一遍反向边就有贡献,就是顺时针和逆时针判断那个最少,sum减掉dfs的就是相反时针 

B题_第1张图片 

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

#define sfi(i) scanf("%d",&i)
#define sfs(i) scanf("%s",(i))
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))'9')
    {
        if(ss=='-')f=-1;ss=getchar();
    }
    while(ss>='0'&&ss<='9')
    {
        x=x*10+ss-'0';ss=getchar();
    }    return f*x;
}

ll power(ll x,ll n)
{
    ll ans=1;
    while(n)
    {
        if(n&1) ans=ans*x;
        x=x*x;
        n>>=1;
    }
    return ans;
}

struct E
{
    int v,c;
    int nex;
}edge[maxn];
int head[maxn];
int cnt=0;
void add(int u,int v,int c)
{
    edge[cnt].v=v;
    edge[cnt].c=c;
    edge[cnt].nex=head[u];
    head[u]=cnt++;

    edge[cnt].v=u;
    edge[cnt].c=c;
    edge[cnt].nex=head[v];
    head[v]=cnt++;
}

int ans=0;
bool vis[maxn];
bool f=0;
int n;
int e=0;
void dfs1(int u)
{
    //cout<>n)
    {
        int sum=0;
        cnt=0;
        for(int i=0;i<=n;i++) head[i]=-1;
        mem(vis,0);

        for(int i=0;i>a>>b>>c;
            add(a,b,c);
            sum+=c;
        }
        ans=0;
        dfs1(1);
        //cout<

 

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