1062:昂贵的聘礼（最短路/枚举）

题面链接

题解

1 //最短路径——Dijkstra算法  

2 //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  

3 //从满足改等级范围的结点组成的子图中用Dijkstra来算出最短路径  

4 //小结，通过枚举的方式可以消除一些图与图之间的限制 

C++代码

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define  pi acos(-1.0)
#define  eps 1e-9
#define  fi first
#define  se second
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug         printf("******
")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  name2str(x) #x
#define  fuck(x)     cout<<#x" = "<<x<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)+
#define  FRint(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("data.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
const int maxn = 205;
const int INF = 0x3f3f3f3f;
int n,mp[maxn][maxn], k[maxn], dis[maxn];
bool cha[maxn];
bool vis[maxn];
int l[maxn],v[maxn];
int p[maxn];
int minn;
int dijkstra()
{
    int minn = INF;
    memset(vis,false,sizeof(vis)); //清除所有点的标号 
    for(int i=1;i<=n;i++)
        dis[i] = INF; //设d[1] = 0,其它为 INF 
    dis[1] = 0; //（此处还未加上进入改点的花费）自己换自己为0 
    
    for(int i=1;i<=n;i++) //循环n 次 
    {
        int x, m = INF;
        for(int y=1;y<=n;y++)
        { //在所有未标号且满足等级限制的点中选出 d值最小的点 x 
            if(!vis[y] && dis[y] <= m && cha[y])
                m = dis[x=y];
        }
        vis[x] = true; //标记点x 
        
        for(int y=1;y<=n;y++) // 对于从x出发的所有边 (x, y)更新dist 
        {
            if(cha[y]) //若满足等级限制
                dis[y] = min(dis[y], dis[x]+mp[x][y]);
        }
    }
    
    for(int y=1;y<=n;y++)
    { //对于每个dist[y]还要满足进入改点的花费 
        dis[y] += p[y];
        minn = min(minn,dis[y]);
    }
    
    return minn; //返回最小值 
}

int main() {
    
        int s, e;
        int size ;
        cin >> size >> n;
        
        //scanf ( "%d%d", &s, &e );
        for ( int i = 0; i <= n; i++ )
            for ( int j = 0; j <= n; j++ )
                mp[i][j] = INF;
                
        for(int i = 1;i <= n ; i++){
            int  x;
            cin >> p[i] >> l[i] >> x;
            
            for(int j = 1;j <= x; j++){
                int t , v;
                cin >> t >> v;
                mp[i][t] = v;
            }
        }                
        minn = INF;
        int k = l[1];
            for(int i=0;i<=size;i++) //枚举 
    {
        memset(cha,false,sizeof(cha));
        for(int j=1;j<=n;j++)
        { //枚举等级允许的范围 
            if(l[j] >=k-size+i && l[j]<=k+i)
                cha[j] = true;
        }
        
        minn = min(minn, dijkstra());
    }
  
      cout << minn << endl;
    
    return 0;
}