UVa 12186 Another Crisis dp：树上dp

UVa 12186 Another Crisis 树上dp

The question is from here.

My Solution

d(i)表示给上级发信至少需要多少工人。把所有子结点的d值排序然后取前c个，

对除非的处理 ★★ c = (k*T - 1) /100 +1;

用前向星储存树  每组数据搞完要 for 1 -> n 对sons[ i ].clear();

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 100008;
int n,t,T;
vector sons[maxn];      //sons[i] 为结点i的子列表

int dp(int u)
{
    if(sons[u].empty()) return 1;
    int k = sons[u].size();
    vector d;
    for(int i = 0; i < k; i++)
        d.push_back(dp(sons[u][i]));
    sort(d.begin(),d.end());
    int c = (k*T-1) /100 +1;
    int ans = 0;
    for(int i = 0; i < c; i++) ans += d[i];
    return ans;
}

int main()
{
    while(scanf("%d%d", &n, &T)){
        if(n == 0 && T == 0) break;
        for(int i = 1; i <= n; i++){
            scanf("%d", &t);
            sons[t].push_back(i);
        }
        printf("%dn", dp(0));
        for(int i = 0; i <= n; i++){
            sons[i].clear();
        }
    }
    return 0;
}

也可以写成这样

#include 
#include 
#include 
#include 
//#define LOCAL
using namespace std;
const int maxn = 100008;
int n,t,T;
vector sons[maxn];      //sons[i] 为结点i的子列表

void build()
{
    for(int i = 1; i <= n; i++){
        scanf("%d", &t);
        sons[t].push_back(i);
    }
}

int dp(int u)
{
    if(sons[u].empty()) return 1;
    int k = sons[u].size();
    vector d;
    for(int i = 0; i < k; i++)
        d.push_back(dp(sons[u][i]));
    sort(d.begin(),d.end());
    int c = (k*T-1) /100 +1;
    int ans = 0;
    for(int i = 0; i < c; i++) ans += d[i];
    return ans;
}

void initial()
{
    for(int i = 0; i <= n; i++){   //清空的时候要把大boss的也清理掉
        sons[i].clear();
    }
}

int main()
{
    #ifdef LOCAL
    freopen("a.txt","r",stdin);
    #endif // LOCAL
    while(scanf("%d%d", &n, &T)){
        if(n == 0 && T == 0) break;
        build();
        printf("%dn", dp(0));
        initial();
    }
    return 0;
}

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