poj 1062 昂贵的聘礼(dfs+剪枝)

http://poj.org/problem?id=1062

　　网上都说是dij，可是我想不到怎么dij。。。- -所以就只好暴力dfs+剪枝来做这题了。实践过，也就47ms，不会太长时间。这题注意的是，最高等级和最低等级的差不超过给定的值的判断。暴力搜索加上剪枝，剪枝的方法只需（1）跟目前找到的最小价格比较，判断是否继续找下去（2）判断是否曾经搜索了。

通过代码：

View Code

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <iomanip>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <string>
 9 #include <queue>
10 #include <ctime>
11 
12 using namespace std;
13 
14 #define PB push_back
15 #define FI first
16 #define SE second
17 #define MPR make_pair
18 #define REP(i, n) for (int i = 0; i < n; i++)
19 #define REP_1(i, n) for (int i = 1; i <= n; i++)
20 #define FORI(i, a, b) for (int i = a; i < b; i++)
21 #define FORD(i, a, b) for (int i = a; i > b; i--)
22 #define _clr(x) memset(x, 0, sizeof(x))
23 #define _rst(x) memset(x, -1, sizeof(x))
24 
25 typedef long long LL;
26 typedef pair<int, int> PII;
27 typedef vector<LL> VLL;
28 typedef vector<PII> VPII;
29 typedef vector<int> VI;
30 typedef vector<double> VDBL;
31 const int N = 105;
32 const int hashMod = 1e6 + 5;
33 const int inf = 0x55555555;
34 const double eps = 1e-8;
35 const LL linf = 0x5555555555555555ll;
36 const double finf = 1e50;
37 const double pi = acos(-1.0);
38 
39 LL minCost;
40 int level;
41 bool vis[N];
42 struct Good {
43     int price, level;
44     VPII sub;
45     Good(int _p = 0, int _l = 0) {
46         price = _p, level = _l;
47         sub.clear();
48     }
49 } good[N];
50 
51 void input(int n) {
52     int p, l, m;
53     REP_1(i, n) {
54         cin >> p >> l >> m;
55         good[i] = Good(p, l);
56         REP(j, m) {
57             cin >> p >> l;
58             good[i].sub.PB(MPR(p, l));
59         }
60     }
61     minCost = finf;
62     _clr(vis);
63 }
64 
65 void dfs(LL curCost, int cur, int mn, int mx) {
66     int sz = good[cur].sub.size();
67     vis[cur] = true;
68     minCost = min(curCost + good[cur].price, minCost);
69     REP(i, sz) {
70         int id = good[cur].sub[i].FI, cost = good[cur].sub[i].SE;
71         if (!vis[id] && max(mx, good[id].level) - min(mn, good[id].level) <= level) {
72             if (curCost + cost <= minCost) {
73                 dfs(curCost + cost, id, min(mn, good[id].level), max(mx, good[id].level));
74             }
75         }
76     }
77     vis[cur] = false;
78 }
79 
80 int main() {
81     int n;
82     while (cin >> level >> n) {
83         input(n);
84         dfs(0, 1, good[1].level, good[1].level);
85         cout << minCost << endl;
86     }
87     return 0;
88 }

——written by Lyon