FAFUOJ 1572 Big castle（樹形DP）

FAFUOJ 1572 Big castle（樹形DP）

題意：給定一棵樹，然後每次可以操作節點，使得節點和周圍節點的狀態都翻轉，問是否能使得所有節點都為1

思路：樹形DP， dp[n][2][2] 的狀態，

#include 
#include 
#include 
#include 
using namespace std;

const int N = 50005;

int n;

int node[N];
bool dp[N][2][2];

void scanf_(int &num)
{
    char in;
    bool neg=false;
    while(((in=getchar()) > '9' || in<'0') && in!='-') ;
    if(in=='-')
    {
        neg=true;
        while((in=getchar()) >'9' || in<'0');
    }
    num=in-'0';
    while(in=getchar(),in>='0'&&in<='9')
        num*=10,num+=in-'0';
    if(neg)
        num=0-num;
}

struct Edge {
    int u, v;
    Edge() {}
    Edge(int u, int v) {
        this->u = u;
        this->v = v;
    }
} edge[N * 2];

int en = 0;
int first[N], nex[N * 2];

void add_edge(int u, int v) {
    edge[en] = Edge(u, v);
    nex[en] = first[u];
    first[u] = en++;
}

void dfs(int u, int p) {
    memset(dp[u], false, sizeof(dp[u]));
    int odd1 = 0, oe1 = 0, even1 = 0, odd2 = 0, oe2 = 0, even2 = 0;
    int sum = 0;
    for (int i = first[u]; i + 1; i = nex[i]) {
        int v = edge[i].v;
        if (v == p) continue;
        dfs(v, u);
        if (dp[v][0][1] && dp[v][0][0]) oe1++;
        else if (dp[v][0][1]) odd1++;
        else if (dp[v][0][0]) even1++;
        if (dp[v][1][1] && dp[v][1][0]) oe2++;
        else if (dp[v][1][1]) odd2++;
        else if (dp[v][1][0]) even2++;
        sum++;
    }
    if (oe2 + odd2 + even2 == sum) {
        if (oe2) dp[u][0][0] = dp[u][1][0] = true;
        else {
            if (odd2&1) dp[u][!node[u]][0] = true;
            else dp[u][node[u]][0] = true;
        }
    }
    if (oe1 + odd1 + even1 == sum) {
        if (oe1) dp[u][0][1] = dp[u][1][1] = true;
        else {
            if (odd1&1) dp[u][node[u]][1] = true;
            else dp[u][!node[u]][1] = true;
        }
    }
}

int main() {
    while (~scanf("%d", &n)) {
        memset(dp, false, sizeof(dp));
        memset(first, -1, sizeof(first));
        en = 0;
        for (int i = 1; i <= n; i++)
            scanf_(node[i]);
        int u, v;
        for (int i = 1; i <= n - 1; i++) {
            scanf_(u);
            scanf_(v);
            add_edge(u, v);
            add_edge(v, u);
        }
        dfs(1, 0);
        if (dp[1][1][0] || dp[1][1][1]) printf("Great Cdfpysw!\n");
        else printf("Poor Nanaya!\n");
    }
    return 0;
}