牛客小白月赛3---G 旅游（树形dp）

Posted 2020-11-04 SomnusMistletoe

题目链接：https://www.nowcoder.com/acm/contest/87/G

分析：

1、对于点cur，dp[cur][0]表示在该点住宿；dp[cur][1]表示其某个子结点住宿，自己被访问到；dp[cur][2]表示其父结点住宿，自己被访问到；

2、如果dp[cur][0]，则cur的子结点全是2状态

3、如果dp[cur][2]，则cur的子结点要么为0状态，要么为1状态，取这两个状态的最大值累加。

4、如果dp[cur][1]，则cur的子结点至少有一个为0状态，其余的为1状态，同样是取两个状态的最大值累加；

不过，若每个子结点都是取1状态时为最优，则必须在他们当中找出一个损失最小的0状态，所以记录每个结点1状态与0状态差值的最小值，然后将变为0状态后损失最小的结点变为0状态即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<fstream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const double pi = acos(-1.0);
const int MAXN = 500000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int dp[MAXN][3];
vector<int> v[MAXN];
int dfs(int cur, int id, int fa){
    if(dp[cur][id] != -INT_INF) return dp[cur][id];
    int len = v[cur].size();
    if(id == 0){
        dp[cur][id] = 1;
        for(int i = 0; i < len; ++i){
            if(v[cur][i] == fa) continue;
            dp[cur][id] += dfs(v[cur][i], 2, cur);
        }
    }
    else if(id == 1 && v[cur].size() > 0){
        dp[cur][id] = 0;
        int dif = INT_INF;
        bool ok = false;
        for(int i = 0; i < len; ++i){
            if(v[cur][i] == fa) continue;
            if(dfs(v[cur][i], 0, cur) >= dfs(v[cur][i], 1, cur)){
                ok = true;
                dp[cur][id] += dfs(v[cur][i], 0, cur);
            }
            else{
                dp[cur][id] += dfs(v[cur][i], 1, cur);
                dif = min(dif, dfs(v[cur][i], 1, cur) - dfs(v[cur][i], 0, cur));
            }
        }
        if(!ok){
            dp[cur][id] -= dif;
        }
    }
    else if(id == 2){
        dp[cur][id] = 0;
        for(int i = 0; i < len; ++i){
            if(v[cur][i] == fa) continue;
            dp[cur][id] += max(dfs(v[cur][i], 0, cur), dfs(v[cur][i], 1, cur));
        }
    }
    return dp[cur][id];
}
int main(){
    int n, s;
    scanf("%d%d", &n, &s);
    for(int i = 1; i <= n; ++i){
        dp[i][0] = dp[i][1] = dp[i][2] = -INT_INF;
    }
    int x, y;
    for(int i = 0; i < n - 1; ++i){
        scanf("%d%d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    printf("%d\n", dfs(s, 0, -1));
    return 0;
}