HDU 4274 Spy's Work (树形DP)

题意

给定一棵树，给出一些子树的权值关系，问是否矛盾（初始所有结点的下限为1）

思路

设lmin和lmax表示题目给定的限制范围，默认为[1..oo]；amin和amax表示实际符合要求的范围。从根节点开始DP，通过子树的amin更新父节点的amin（父节点的amax一定是oo，因为它自身权值任意），判断此时amin和amax与lmin和lmax是否有交集（实际范围），没有则false。

代码

[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
#define REP(i, begin, end) for (int i = begin; i <= end; i ++)
using namespace std;

const int N = 10005;
const int oo = 0x7fffffff;
long long lmin
, lmax
, amin
, amax
;
vector <int> adj
;

bool dfs(int u){
bool ok = true;
for (int i = 0; i < (int)adj[u].size(); i ++){
int v = adj[u][i];
ok = dfs(v);
if (ok == false)
return false;
amin[u] += amin[v];
amax[u] += amax[v];
}
amin[u] = max(amin[u], lmin[u]);
amax[u] = min(amax[u], lmax[u]);
if (amin[u] <= amax[u]) return true;
else return false;
}
int main(){
int n;
while(scanf("%d", &n) != EOF){
for (int i = 1; i <= n; i ++){
adj[i].clear();
lmin[i] = 1;
lmax[i] = oo;
amin[i] = 1;
amax[i] = oo;
}
for (int i = 2; i <= n; i ++){
int tmp;
scanf("%d", &tmp);
adj[tmp].push_back(i);
}
int q;
scanf("%d", &q);
bool res = true;
for (int i = 0; i < q; i ++){
int a, b;
char c;
scanf("%d%*c%c%*c%d", &a, &c, &b);
if (c == '='){
if (b <= lmax[a] && b >= lmin[a]) lmin[a] = lmax[a] = b;
else res = false;
}
else if (c == '<'){
lmax[a] = min((long long)(b - 1), lmax[a]);
}
else if (c == '>'){
lmin[a] = max((long long)(b + 1), lmin[a]);
}
}
if (!res){
puts("Lie");
}
else{
for (int i = 1; i <= n; i ++){
if (lmin[i] > lmax[i]){
res = false;
break;
}
}
if (!res){
puts("Lie");
}
else{
res = dfs(1);
if (res){
puts("True");
}
else{
puts("Lie");
}
}
}
}
return 0;
}
[/cpp]