如题，openjudge的代码可以过洛谷的，但是洛谷能过的过不了openjudge 这是洛谷代码

#include <bits/stdc++.h>
using namespace std;
#define limit (1000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 988244353
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n,m,vs,ve;
int head[limit],cnt;
int dist[limit],fa[limit],vis[limit],pre[limit];
struct node{
    int to, next, flow, w;
}edge[limit<<1];
void add(int u, int v, int flow, int w = 1){
    edge[cnt].to = v;
    edge[cnt].flow = flow;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
void init(int flag = 0){
    if(flag){
        memset(dist, INF, sizeof(dist));
        memset(vis, 0 , sizeof(vis));
    }else{
        memset(head, -1, sizeof(head));
        cnt = 0;
    }
}
int spfa(){
    queue<int>q;
    init(1);
    dist[vs] = 0;
    vis[vs] = 1;
    q.push(vs);
    pre[ve] = -1;
    while (q.size()){
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = edge[i].next){
            int v = edge[i].to, w = edge[i].w;
            int stream = edge[i].flow;
            if(dist[u] + w < dist[v] && stream > 0){
                dist[v] = dist[u] + w;
                pre[v] = i;
                fa[v] = u;
                if(!vis[v]){
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    return ~pre[ve];
}

int max_flow,min_cost;
void dinic(){
    while (spfa()){
        int min_flow = INF;
        for(int i = ve; i != vs; i = fa[i]){
            min_flow = min(min_flow, edge[pre[i]].flow);
        }
        max_flow += min_flow;
        min_cost  += dist[ve] * min_flow;
        for(int i = ve; i != vs; i = fa[i]){
            edge[pre[i]].flow -= min_flow;
            edge[pre[i] ^ 1].flow += min_flow;
        }
    }
}
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    n = read(), m = read(), vs= 1, ve =m;
    init();
    rep(i, 1, n){
        int x= read(),y = read(), floww = read();
        if(~head[x] || edge[head[x]].flow == 0) {
            add(x, y, floww);
            add(y, x, 0);
        }else{
            edge[head[x]].flow += floww;
        }
    }
    dinic();
    printf("%d", max_flow);
    return 0;
}

这是openjudge代码：

int G[limit][limit];
int s,e,f;//分别表示起点，终点和流量
int visited[limit], level[limit];
bool countLayer(){
    int layer = 0;
    deque<int>q;//当做栈使用的q
    memset(level, 0xff , sizeof(level));//初始化为-1表示没有遍历过
    level[1] = 0;//表示初始分层
    q.push_back(1);//推入栈中以开始编号
    while(!q.empty()){
        int vertices = q.front();
        q.pop_front();
        for(int i = 1 ; i <= e; ++i){
            if(G[vertices][i] > 0 && level[i] == -1){
                
                level[i] = level[vertices] + 1;
                if(i == e){
                    
                    return true;
                }else{
                    q.push_back(i);//继续explore下一个节点
                }
            }
        }
    }
    return false;
}
int dinic(){
    int i, start, maxFlow = 0;
    
    deque<int>q;
    
    while(countLayer()){
        q.push_back(1);
        memset(visited , 0, sizeof(visited));
        visited[1] = 1;
        while(!q.empty()){
            int n = q.back();
            if(n == e){
                
                int minC = 999999;
                int minV;//minV记录最小流出现的节点
                for(i = 1 ; i < q.size(); i++){
                    int vs = q[i - 1];//记录起始点
                    int ve = q[i];//记录终点
                    if(G[vs][ve] > 0){
                        if(minC > G[vs][ve]){
                            minC = G[vs][ve];
                            minV = vs;
                        }
                    }
                }
                maxFlow += minC;
                for(i = 1 ; i < q.size();++i){
                    int vs = q[i - 1];//记录起始点
                    int ve = q[i];//记录终点
                    G[vs][ve] -= minC;
                    G[ve][vs] += minC;
                }
                while(!q.empty() && q.back() != minV){
                    visited[q.back()] = 0;
                    q.pop_back();
                }
            }else{
                for(i = 1 ; i <= e ; ++i){
                    if(G[n][i] > 0 && level[i] == level[n] + 1 && !visited[i]){
                        
                        visited[i] = 1;
                        break;
                    }

                }
                if(i > e){
                    q.pop_back();，
                    
                }
            }
        }
    }
    return maxFlow;
}
int main(){
    while(scanf("%d%d", &s, &e) == 2){
        memset(G,0, sizeof(G));//初始化数组存边
        int vs, ve ,flow;
        for(int i = 0 ; i < s; ++i){
            scanf("%d%d%d" , &vs, &ve,&flow);
            G[vs][ve] += flow;
        }
        printf("%d\n", dinic());//大功告成啦！
    }
    return 0;
}