例题9-12 工人的请愿书 UVa12186

1.题目描述：点击打开链接

2.解题思路：本题利用树状dp解决，不过其实也可以理解为用贪心法解决的。设d(u)表示u给上级发信最少需要的工人个数，假设u有k个子结点，那么根据题意，至少需要c=(k*T-1)/100+1个直属下属发信才行。而每个直属下属的工人数是di，那么这时只需要把di由小到大排序，然后把前c个相加就是d(u)了。最终的答案是d(0)。由于需要排序，因此总的时间复杂度是O(N*logN)。最后附上一篇带有大量宏定义的参考代码，真心给跪，第一次见这么多的宏定义，Orz。

3.代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

const int maxn = 1e5 + 10;
vector<int>sons[maxn];//存放i的所有直属下属
int n, T;
int dp(int u)
{
	if (sons[u].empty())return 1;//工人
	int k = sons[u].size();
	vector<int>d; //存放u的直属下属的工人数
	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()
{
	//freopen("test.txt", "r", stdin);
	while (scanf("%d%d", &n, &T) == 2 && n&&T)
	{
		memset(sons, 0, sizeof(sons));
		int x;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			sons[x].push_back(i);
		}
		int ans;
		ans = dp(0);
		printf("%d\n", ans);
	}
	return 0;
}

参考代码：

 

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <queue>
#include <cmath>
#include <cstdlib>

using namespace std;

#define PB              push_back
#define SIZE(x)         (int)x.size()
#define clr(x,y)        memset(x,y,sizeof(x))
#define RS(n)           scanf ("%s", n)
#define RD(n)           scanf ("%d", &n)
#define RF(n)           scanf ("%lf", &n)
#define ALL(t)          (t).begin(),(t).end()
#define FOR(i,n,m,step) for (int i = n; i <= m; i += step)
#define ROF(i,n,m,step) for (int i = n; i >= m; i -= step)
#define IT              iterator

typedef long long               ll;
typedef unsigned int            uint;
typedef unsigned long long      ull;
typedef vector<int>             vint;
typedef vector<long long>       vll;
typedef vector<string>          vstring;
typedef pair<int, int>          PII;

/*****************************************************************/
const int maxn = 100000 + 5;
vint sons[maxn];
int T;

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

int main() {
    int n;
    while (cin>>n>>T,n) {
        FOR(i,0,n,1) sons[i].clear();
        FOR(i,1,n,1) {
            int tmp;
            RD(tmp);
            sons[tmp].PB(i);
        }
        cout<<dp(0)<<endl;
    }
}