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 }