HDU 4274 Spy's Work （树形DP）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4274

题意：给一颗树，1是根节点，树上每个结点的权值是该节点子树的权值之和（必然不小于1），现在给出某些结点权值的推算关系，问这些关系是否矛盾

思路：对于每个结点，我们可以在dfs时可以通过子树下限更新其权值下限，上限无法确定，然后再根据所给的信息判断是否矛盾

#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <utility>#include <iostream>#include <algorithm>#include <functional>#include <set>#include <map>using namespace std;const int maxn = 10010;const long long inf = 1e17;typedef long long ll;int cnt, head[maxn];int n;struct node{    ll l, r;} a[maxn];struct edge{    int to, nxt;} e[maxn << 1];void init(){    cnt = 0;    memset(head, -1, sizeof(head));    for (int i = 1; i <= n; i++)    {        a[i].l = 1, a[i].r = inf;    }}void add(int u, int v){    e[cnt].to = v;    e[cnt].nxt = head[u];    head[u] = cnt++;}bool dfs(int u, int fa){    long long sl = 0;    for (int i = head[u]; ~i; i = e[i].nxt)    {        int v = e[i].to;        if (fa == v) continue;        if (!dfs(v, u))            return false;        sl += a[v].l;    }    a[u].l = max(a[u].l, sl + 1);    return a[u].l <= a[u].r;}int main(){    while (~scanf("%d", &n))    {        init();        for (int i = 2; i <= n; i++)        {            int u;            scanf("%d", &u);            add(u, i);            add(i, u);        }        int m;        scanf("%d", &m);        while (m--)        {            int u, num;            char s;            scanf("%d %c %d", &u, &s, &num);            if (s == '=')            {                a[u].l = max(a[u].l, (ll)num);                a[u].r = min(a[u].r, (ll)num);            }            if (s == '<')            {                a[u].r = min(a[u].r, (ll)(num - 1));            }            if (s == '>')            {                a[u].l = max(a[u].l, (ll)(num + 1));            }        }        //        for(int i = 1; i <= n; i++)        //            cout << a[i].l << " " <<a[i].r << endl;        if (dfs(1, -1))            puts("True");        else            puts("Lie");    }    return 0;}

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