pku 1052 昂贵的聘礼

题意：中文。。

思路：

这里的等级限制是，与他本身交易的以及间接交易的都不能超过m。所以我们在树形DP时，维护可行区间即可。  关键是在维护可行区间时卡住了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll unsigned __int64
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define N 107

using namespace std;

int len[N][N];

int level[N],val[N];
int dp[N];
int n,m;


int dfs(int u,int Li,int Ri)
{
    int i;
    if (dp[u]) return dp[u];
    dp[u] = val[u];

    for (i = 1; i <= n; ++i)
    {
        if (len[u][i] && level[i] >= Li - m && level[i] <= Ri + m)
        {
            dp[u] = min(dp[u],dfs(i,Li > level[i]?Li:level[i],Ri < level[i]? Ri:level[i]) + len[u][i]);
        }
    }
    return dp[u];
}
int main()
{
//    Read();
    int P,L,X;
    int T,V;
    int i,j;
    while (~scanf("%d%d",&m,&n))
    {
        CL(len,0); CL(dp,0);
        for (i = 1; i <= n; ++i)
        {
            scanf("%d%d%d",&P,&L,&X);
            level[i] = L; val[i] = P;
            for (j = 0; j < X; ++j)
            {
                scanf("%d%d",&T,&V);
                len[i][T] = V;
            }
        }
        printf("%d\n",dfs(1,level[1],level[1]));
    }
    return 0;
}