IOI2008 Island 岛屿

题目描述：

bz

luogu

题解：

裸的基环树直径。

代码：

#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int N = 1000050;
template
inline void read(T&x)
{
    T f = 1,c = 0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();}
    x = f*c;
}
int n,hed[N],cnt=1;
ll ans;
struct EG
{
    int to,nxt;
    ll w;
}e[2*N];
void ae(int f,int t,ll w)
{
    e[++cnt].to = t;
    e[cnt].nxt = hed[f];
    e[cnt].w = w;
    hed[f] = cnt;
}
int sta[2*N],tl,rt;
ll ste[N];
bool vis[N],cir[N],use[N];
int dfs0(int u,int pre)
{
    if(vis[u])
    {
        rt = u;
        return 1;
    }
    vis[u] = 1;
    for(int j=hed[u],now;j;j=e[j].nxt)
    {
        int to = e[j].to;
        if(j==pre)continue;
        if((now=dfs0(to,j^1)))
        {
            if(now==1)
            {
                sta[++tl] = u;
                ste[tl] = e[j].w;
                cir[u] = 1;
                if(u!=rt)return 1;
            }
            return 2;
        }
    }
    return 0;
}
int fa[N];
ll dp[N],sum[2*N],st[2*N],h,l,now;
void bfs(int u)
{
    h = 1,l = 0;
    st[++l] = u;
    use[u]=1;
    while(h<=l)
    {
        u = st[h++];
        for(int j=hed[u];j;j=e[j].nxt)
        {
            int to = e[j].to;
            if(use[to]||cir[to])continue;
            use[to] = 1;
            fa[to] = u;
            st[++l] = to;
        }
    }
    for(int i=l;i>=1;i--)
    {
        u = st[i];
        for(int j=hed[u];j;j=e[j].nxt)
        {
            int to = e[j].to;
            if(cir[to]||to==fa[u])continue;
            ll tmp = dp[to]+e[j].w;
            now = max(now,dp[u]+tmp);
            dp[u]=max(dp[u],tmp);
        }
    }
}
int main()
{
//  freopen("tt.in","r",stdin);
    read(n);
    for(int f,w,i=1;i<=n;i++)
    {
        read(f),read(w);
        ae(f,i,w),ae(i,f,w);
    }
    for(int i=1;i<=n;i++)if(!use[i])
    {
        tl=0;now=0;
        dfs0(i,0);
        for(int j=1;j<=tl;j++)
            bfs(sta[j]),sta[j+tl]=sta[j];
        if(tl==1)
        {
            now = max(now,dp[sta[1]]+ste[1]);
            ans+=now;
            continue;
        }
        for(int j=1;j<=tl;j++)sum[j]=sum[j-1]+ste[j];
        for(int j=1;j<=tl;j++)sum[j+tl]=sum[j+tl-1]+ste[j];
        h = 1,l = 0;
        for(int j=1;j<=2*tl;j++)
        {
            while(h;
            if(h<=l)now = max(now,dp[sta[j]]+sum[j]+dp[sta[st[h]]]-sum[st[h]]);
            while(h;
            st[++l] = j;
        }
        ans+=now;
    }
    printf("%lld\n",ans);
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/LiGuanlin1124/p/10777132.html