BZOJ1304 CQOI2009叶子的染色(树形dp)

  令f[i]表示i子树内最少染色次数,加上012状态分别表示该子树内叶节点已均被满足、存在黑色叶节点未被满足、存在白色叶节点未被满足,考虑i节点涂色情况即可转移。事实上贪心也可以。

  

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 100010
int n,m,f[N][3],c[N],p[N],t=0;
struct data{int to,nxt; 
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dfs(int k,int from)
{
    if (k<=m) f[k][0]=1,f[k][c[k]]=0,f[k][c[k]]=n;
    else
    {
        f[k][0]=f[k][1]=f[k][2]=0;
        for (int i=p[k];i;i=edge[i].nxt)
        if (edge[i].to!=from)
        {
            dfs(edge[i].to,k);
            f[k][0]+=f[edge[i].to][0];
            f[k][1]+=min(f[edge[i].to][1],f[edge[i].to][0]);
            f[k][2]+=min(f[edge[i].to][2],f[edge[i].to][0]);
        }
        f[k][0]=min(f[k][0],min(f[k][1],f[k][2])+1);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj1304.in","r",stdin);
    freopen("bzoj1304.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=m;i++) c[i]=read()+1;
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    dfs(n,n);
    cout<<f[n][0];
    return 0;
}

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转载自www.cnblogs.com/Gloid/p/9544098.html