luogu P2018 消息传递

  一道显而易见的树形dp

每个人都有当根节点的机会

然后先告诉需要花费时间长的，再告诉短的

转移方程

dp[u]=max(dp[u],dp[v]+i−1)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 1010
#define INF 999999999

int n;
int dp[maxn];
int head[maxn],cnt;
int www[maxn],minn = INF;

struct EDGE {
    int nxt,to;
} edge[maxn * 2];

int cmp(int a,int b) {
    return a > b;
}

void add(int x,int y) {
    edge[++cnt].to = y;
    edge[cnt].nxt = head[x];
    head[x] = cnt;
}

void dfs(int u,int fa) {
    int son[maxn] = {0};
    int total = 0;
    for(int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if(v != fa) {
            dfs(v,u);
            son[++total] = dp[v];
        }
    }
    sort(son + 1,son + total + 1,cmp);//先传给花时间长的
    for(int i = 1; i <= total; i++)
        dp[u] = max(dp[u],son[i] + i - 1);//转移方程，儿子到根节点的距离i再减1
    dp[u] += 1;
}

int main() {
    scanf("%d",&n);
    for(int i = 2; i <= n; i++) {
        int a;
        scanf("%d",&a);
        add(a,i);
        add(i,a);
    }
    for(int i = 1; i <= n; i++) {//每个点都当一遍根节点
        memset(dp,0,sizeof(dp));//别忘了每次清零
        dfs(i,0);
        minn = min(minn,dp[i]);存最小值
        www[i] = dp[i];//记录每次的值
    }
    printf("%d
",minn);
    for(int i = 1; i <= n; i++)
        if(minn == www[i])
            printf("%d ",i);
    return 0;
}

集训回来，从今天开始疯狂补博客了

做了蛮多题，都想写写题解加深印象，希望不会鸽