题目：给出一棵树，给出一些子树的权值关系，问是否矛盾
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初始对于所有结点以及子树的下限为1，上限可能不确定
然后通过给出的不等式去更新上下界，之后进行树形DP，通过孩子的下界更新父节点的下界
再判断是否矛盾
注意当前节点当个的权值下限为1，该节点的下限为孩子下界的和+1
[cpp] #include?? #include?? #include?? #include?? #include?? #include?? #include?? #include?? #define inf 110000000?? #define M 10005?? #define N 10005?? #define Min(a,b) ((a)<(b)?(a):(b))?? #define Max(a,b) ((a)>(b)?(a):(b))?? #define pb(a) push_back(a)?? #define mem(a,b) memset(a,b,sizeof(a))?? #define LL long long?? #define MOD 1000000007?? using namespace std;? struct Node{? ??? int v,next;? }edge[N*2];? int n,start[N],tot;? int up[N],low[N];? void addedge(int u,int v){? ??? edge[tot].v=v;? ??? edge[tot].next=start[u];? ??? start[u]=tot++;? }? void _addedge(int u,int v){? ??? addedge(u,v);? ??? addedge(v,u);? }? bool ans;? void dfs(int u,int pre){? ??? if(ans) return;? ??? if(up[u]!=-1&&low[u]>up[u]){ans=true;return;}? ??? int tmp=0;? ??? int leaf=1;? ??? for(int i=start[u];i!=-1;i=edge[i].next){? ??????? int v=edge[i].v;? ??????? if(v==pre) continue;? ??????? dfs(v,u);? ??????? leaf=0;? ??????? tmp+=low[v];? ??? }? ??? if(leaf) return;? ??? low[u]=max(tmp+1,low[u]);? ??? if(up[u]!=-1&&low[u]>up[u]) ans=true;? }? int main(){? ??? while(scanf("%d",&n)!=EOF){? ??????? tot=0;mem(start,-1);? ??????? for(int i=2;i<=n;i++){? ??????????? int u;? ??????????? scanf("%d",&u);? ??????????? _addedge(u,i);? ??????? }? ??????? int m;? ??????? for(int i=1;i<=n;i++){low[i]=1;up[i]=-1;}? ??????? scanf("%d",&m);? ??????? for(int i=0;i ??????????? int u,w;? ??????????? char str[5];? ??????????? scanf("%d%s%d",&u,str,&w);? ??????????? if(str[0]=='<') up[u]=w-1;? ??????????? else if(str[0]=='>') low[u]=w+1;? ??????????? else low[u]=up[u]=w;? ??????? }? ??????? ans=false;? ??????? dfs(1,0);? ??????? puts(ans?"Lie":"True");? ??? }? ??? return 0;? }?