傳送門
先考慮什么時候不合法。
第一是考慮任意兩個特殊點的權值的奇偶性是否滿足條件。
第二是考慮每個點的取值范圍是否合法。
如果上述條件都滿足的話就可以隨便構造出一組解。
代碼：
#include<bits/stdc++.h>
#define N 100005
#define inf 0x3f3f3f3f
using namespace std
;
inline int read(){int ans
=0;char ch
=getchar();while(!isdigit(ch
))ch
=getchar();while(isdigit(ch
))ans
=(ans
<<3)+(ans
<<1)+(ch
^48),ch
=getchar();return ans
;
}
int n
,k
,first
[N
],cnt
,fa
[N
],val
[N
],low
[N
],high
[N
],ans
[N
],rt
;
bool is
[N
];
struct edge
{int v
,next
;}e
[N
<<1];
inline void add(int u
,int v
){e
[++cnt
].v
=v
,e
[cnt
].next
=first
[u
],first
[u
]=cnt
;}
inline int max(int a
,int b
){return a
>b
?a
:b
;}
inline int min(int a
,int b
){return a
<b
?a
:b
;}
inline bool dfs1(int p
,int fa
,bool tmp
){if(is
[p
]&&tmp
!=(val
[p
]&1))return false;for(int i
=first
[p
];i
;i
=e
[i
].next
){int v
=e
[i
].v
;if(v
==fa
)continue;if(!dfs1(v
,p
,(tmp
^1)))return false;}return true;
}
inline bool dfs2(int p
,int fa
){if(is
[p
])low
[p
]=high
[p
]=val
[p
];for(int i
=first
[p
];i
;i
=e
[i
].next
){int v
=e
[i
].v
;if(v
==fa
)continue;if(!dfs2(v
,p
))return false;low
[p
]=max(low
[p
],low
[v
]-1),high
[p
]=min(high
[p
],high
[v
]+1);}return low
[p
]<=high
[p
];
}
inline void dfs3(int p
,int fa
,int w
){ans
[p
]=w
;for(int i
=first
[p
];i
;i
=e
[i
].next
){int v
=e
[i
].v
;if(v
==fa
)continue;dfs3(v
,p
,(low
[v
]<=w
-1&&high
[v
]>=w
-1?w
-1:w
+1));}
}
int main(){n
=read();for(int i
=1;i
<n
;++i
){int a
=read(),b
=read();add(a
,b
),add(b
,a
);}for(int i
=1;i
<=n
;++i
)low
[i
]=-inf
,high
[i
]=inf
;k
=read();for(int i
=1;i
<=k
;++i
){int a
=read(),b
=read();val
[a
]=b
,is
[a
]=1,rt
=a
;}if(!dfs1(rt
,0,(val
[rt
]&1))||!dfs2(rt
,0)){puts("No");return 0;}puts("Yes");dfs3(rt
,0,val
[rt
]);for(int i
=1;i
<=n
;++i
)printf("%d\n",ans
[i
]);return 0;
}
轉載于:https://www.cnblogs.com/ldxcaicai/p/9738231.html
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