poj 1062（最短路）

此题的关键在于等级限制的处理，最好的办法是采用枚举，即假设酋长等级为5，等级限制为2，那么需要枚举等级从3~5,4~6,5~7.

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<cmath>
  6 #include<queue>
  7 #include<stack>
  8 #include<string>
  9 #include<vector>
 10 #include<cstdlib>
 11 #include<map>
 12 #include<set>
 13 using namespace std;
 14 #define CL(x,v) memset(x,v,sizeof(x));
 15 #define R(i,st,en) for(int i=st;i<en;++i)
 16 #define LL long long
 17 #define inf 0x3f3f3f3f
 18 
 19 const int maxn = 105;
 20 int m, n;
 21 struct node
 22 {
 23     int p;
 24     int le;
 25 }goods[maxn];
 26 struct edge
 27 {
 28     int v;
 29     int c;
 30     edge(){}
 31     edge(int v, int c):v(v),c(c){}
 32 };
 33 vector <edge> adj[maxn];
 34 int LP,maxLe;
 35 bool vis[maxn];
 36 int dis[maxn];
 37 int dij()
 38 {
 39     CL(dis,0x37);
 40     dis[1] = 0;
 41     for (int i = 0; i < adj[1].size(); ++ i)
 42     {
 43         if(!vis[adj[1][i].v])
 44             dis[adj[1][i].v] = adj[1][i].c;
 45     }
 46     for (int i = 1 ; i < n; ++ i)
 47     {
 48         int tmp = inf, k = 1;
 49         for (int j = 1; j <= n; ++ j)
 50         {
 51             if(!vis[j] && (tmp > dis[j]))
 52             {
 53                 tmp = dis[j];
 54                 k = j;
 55             }
 56         }
 57         vis[k] = 1;
 58         for (int j = 0; j < adj[k].size(); ++ j)
 59         {
 60             if(!vis[adj[k][j].v])
 61                 dis[adj[k][j].v] = min(dis[adj[k][j].v], dis[k] + adj[k][j].c);
 62         }
 63     }
 64     int minans = inf;
 65     for (int i = 1; i <= n; ++ i)
 66     {
 67         if(minans > dis[i] + goods[i].p)
 68             minans = dis[i] + goods[i].p;
 69     }
 70     return minans;
 71 }
 72 int main()
 73 {
 74     while(~scanf("%d%d",&m, &n))
 75     {
 76         CL(goods,0);
 77         CL(adj,0);
 78         maxLe = 0;
 79         R(i, 1, n+1)
 80         {
 81             int x, v, c;
 82             scanf("%d%d%d", &goods[i].p, &goods[i].le, &x);
 83             if(goods[i].le > maxLe)
 84                 maxLe = goods[i].le;
 85             while(x --)
 86             {
 87                 scanf("%d%d", &v, &c);
 88                 adj[i].push_back(edge(v,c));
 89             }
 90         }
 91         LP = goods[1].le;
 92         int st = (LP - m) > 0 ? (LP - m) : 0;
 93         int ans = inf;
 94         for (int i = st; i <= LP; ++ i)
 95         {
 96             CL(vis,0);
 97             for (int j = 1; j <= n; ++ j)
 98             {
 99                 if(goods[j].le < i || abs(goods[j].le - i) > m)
100                     vis[j] = 1;
101             }
102             ans = min(ans, dij());
103         }
104         printf("%d\n",ans);
105     }
106     return 0;
107 }

 

 

posted @ 2013-04-02 10:26  Missa  阅读(216)  评论(0)  编辑  收藏  举报