一般求最短路，限制某个条件

Gym 102501A Environment-Friendly Gym 102501A Environment-Friendly

题意：求最小的co2消耗量（最短路可） ，有一个限制条件，路途的距离 不能超过B
思路：dj+dp
代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define INF 0x3f3f3f3f3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define se second
#define MA(a, b) make_pair(a, b)
using namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<int,int> pii;
int dx[4]= {-1,1,0,0},dy[4]= {0,0,1,-1};
const ll mod=10001;
const ll N =1e3+10;
const double pi=acos(-1);
const double eps = 1e-4;
int h[N*N],e[N*N],ne[N*N],idx,w[N*N];
int B;
int n;
struct piont
{int x,y;int dis(piont a){return ceil(sqrt((x-a.x)*(x-a.x)+(y-a.y)*(y-a.y)));}
}a[N];
struct node
{int x,y,w;bool operator <(const node &M)const{return w>M.w;}
};
int c[110];
int dis1[110][N];
int vis[110][N];
void add(int a,int b,int w1)
{w[idx]=w1,e[idx]=b,ne[idx]=h[a];h[a]=idx++;
}
void dj()
{priority_queue<node>q;for(int i=0;i<=B;i++){for(int j=0;j<n;j++)dis1[i][j]=0x3f3f3f3f;dis1[i][1001]=0x3f3f3f3f;dis1[i][1002]=0x3f3f3f3f;}dis1[0][1001]=0;q.push({0,1001,0});while(q.size()){node t=q.top();q.pop();if(vis[t.x][t.y]) continue;vis[t.x][t.y]=1;for(int i=h[t.y];i!=-1;i=ne[i]){int v1=e[i];int ch=a[t.y].dis(a[v1]);int w1=w[i]*ch;if(t.x+ch<=B&&dis1[t.x][t.y]+w1<dis1[t.x+ch][v1]){dis1[t.x+ch][v1]=dis1[t.x][t.y]+w1;q.push({t.x+ch,v1,dis1[t.x+ch][v1]});}}
}}
int main()
{FILL(h,-1);scanf("%d%d%d%d%d%d",&a[1001].x,&a[1001].y,&a[1002].x,&a[1002].y,&B,&c[0]);int T;scanf("%d",&T);rep1(i,T) scanf("%d",&c[i]);scanf("%d",&n);rep(i,n){int x;scanf("%d%d%d",&a[i].x,&a[i].y,&x);rep1(j,x){int v,j1;scanf("%d%d",&v,&j1);add(i,v,c[j1]);add(v,i,c[j1]);}}rep(i,n){add(i,1001,c[0]);add(1001,i,c[0]);add(i,1002,c[0]);add(1002,i,c[0]);}add(1002,1001,c[0]);add(1001,1002,c[0]);dj();int any=0x3f3f3f3f;for(int i=0;i<=B;i++) any=min(any,dis1[i][1002]);if(any==0x3f3f3f3f) printf("-1");else printf("%d",any);return 0;
}