DongDong fly

Topic connection: https://ac.nowcoder.com/acm/contest/904/D

　　The first study about this type of exercise, it is quite easy to understand, because the limited number of half-price, so in each case should be written every one, dp [Position] [number] pushed to dp [arrival location] [times] and dp [reach the position] [number of + 1] in both cases, and then click the shortest run on it

　　AC Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef struct W_W{
    int eend;
    int weight;
    int next;
}miao;
typedef struct W_w{
    int eend;
    int ci;
}wang;
ll minn(ll a,ll b){
    if(a<b) return a;
    return b;
}
miao x[100010];
int head[100010];
ll dp[100010][15];
int vis[100010][15];
int cnt=0;
void add(int a,int b,int c){
    x[cnt].eend=b;
    x[cnt].weight=c;
    x[cnt].next=head[a];
    head[a]=cnt++;
}
ll spfa(int start,int eend,int k){
    queue<wang> q1;
    q1.push({1,0});
    dp[1][0]=0;
    while(q1.size()){
        int dang=q1.front().eend;
        int ci=q1.front().ci;
        //printf("+++%d %d %lld \n",dang,ci,dp[dang][ci]);
        q1.pop();
        vis[dang][ci]=0;
        for(int i=head[dang];i!=-1;i=x[i].next){
            int to=x[i].eend;
            if(dp[to][ci]==-1){
                dp[to][ci]=dp[dang][ci]+x[i].weight;
                if(vis[to][ci]==0){
                    q1.push({to,ci});
                    vis[to][ci]=1;
                }
            }
            else{
                if(dp[to][ci]>dp[dang][ci]+x[i].weight){
                    dp[to][ci]=dp[dang][ci]+x[i].weight;
                    if(vis[to][ci]==0){
                        q1.push({to,ci});
                        vis[to][ci]=1;
                    }
                }
            }
            if(ci+1<=k){
                if(dp[to][ci+1]==-1){
                    dp[to][ci+1]=dp[dang][ci]+x[i].weight/2;
                    if(vis[to][ci+1]==0){
                        q1.push({to,ci+1});
                        vis[to][ci+1]=1;
                    }
                }
                else{
                    if(dp[to][ci+1]>dp[dang][ci]+x[i].weight/2){
                        dp[to][ci+1]=dp[dang][ci]+x[i].weight/2;
                        if(vis[to][ci+1]==0){
                            q1.push({to,ci+1});
                            vis[to][ci+1]=1;
                        }
                    }
                }
            }
        }
    }
//    for(int i=1;i<=eend;i++){
//        for(int j=0;j<=k;j++){
//            printf("(%d %d) %lld ",i,j,dp[i][j]);
//        }
//        printf("\n");
//    }
    ll ans=-1;
    for(int i=0;i<=k;i++){
       // printf("+++%lld %d %d %lld\n",ans,eend,i,dp[eend][i]);
        if(dp[eend][i]!=-1){
            if(ans==-1){
                ans=dp[eend][i];
            }
            else{
                ans=minn(dp[eend][i],ans);
            }
        }
    }
    return ans;
}
int main()
{
    int m,n,k;
    scanf("%d %d %d",&m,&n,&k);
    memset(head,-1,sizeof(head));
    memset(dp,-1,sizeof(dp));
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n;i++){
        int a,b;
        int c;
        scanf("%d %d %d",&a,&b,&c);
        add(a,b,c);
    }
    ll ans=spfa(1,m,k);
    printf("%lld\n",ans);
    return 0;
}