CF360E Levko and Game (greedy)

This problem greedy without water, looking for \ (dis1 <= dis2 \) points to hysterically cut, Kandao is not cut.
Write blog is to record the magic bug encountered under

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))

#define ON_DEBUGG

#ifdef ON_DEBUGG

#define D_e_Line printf("\n-----------\n")
#define D_e(x) std::cout << (#x) << " : " <<x << "\n"
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#define Pause() system("pause")
#include <ctime>
#define TIME() fprintf(stderr, "\nTIME :　%.3lfms\n", clock() * 1000.0 / CLOCKS_PER_SEC)

#else

#define D_e_Line ;
#define D_e(x) ;
#define FileOpen() ;
#define FilSave ;
#define Pause() ;
#define TIME() ;

#endif

struct ios {
    template<typename ATP> ios& operator >> (ATP &x) {
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x *= f;
        return *this;
    }
}io;

using namespace std;

template<typename ATP> inline ATP Min(ATP a, ATP b) {
    return a < b ? a : b;
}
template<typename ATP> inline ATP Max(ATP a, ATP b) {
    return a > b ? a : b;
}
#include <queue>
const int N = 100007;
const int M = 1000007;

#define int long long
struct Edge {
    int nxt, pre, from, w;
} e[M];
int head[N], cntEdge;
inline void add(int u, int v, int w) {
    e[++cntEdge] = (Edge){ head[u], v, u, w}, head[u] = cntEdge;
}

struct nod {
    int x, w;
    bool operator < (const nod &com) const {
        return w > com.w;
    }
};
priority_queue<nod> q;
int dis1[N], dis2[N];
int n, m, K, S1, S2, T;
inline void Dijkstra(int st, int *dis) {
//  Fill(dis, 0x3f); This sentence lead to a bug
/*
[Warning] argument to 'sizeof' in 'void* memset(void*, int, size_t)' call is 
the same expression as the destination; did you mean to dereference it? [-Wsizeof-pointer-memaccess]
*/
    R(i,0,n) dis[i] = 0x7fffffff;
    dis[st] = 0;
    q.push((nod){ st, 0});
    while(!q.empty()){
        int u = q.top().x, w = q.top().w;
        q.pop();
        if(w != dis[u]) continue;
        for(register int i = head[u]; i; i = e[i].nxt){
            int v = e[i].pre;
            if(dis[v] > dis[u] + e[i].w){
                dis[v] = dis[u] + e[i].w;
                q.push((nod){ v, dis[v]});
            }
        }
    }
}

int l[N], r[N], pos[N];
#undef int
int main() {
#define int long long
    io >> n >> m >> K >> S1 >> S2 >> T;
    R(i,1,m){
        int u, v, w;
        io >> u >> v >> w;
        add(u, v, w);
    }
    R(i,1,K){
        int u, v;
        io >> u >> v;
        io >> l[i] >> r[i];
        add(u, v, r[i]);
        pos[i] = cntEdge;
    }
    while(1){
        int flag = false;
        Dijkstra(S1, dis1);
        Dijkstra(S2, dis2);
        R(i,1,K){
            int v = e[pos[i]].from;
            if((dis1[v] <= dis2[v]) && e[pos[i]].w != l[i]){
                e[pos[i]].w = l[i];
                flag = true;
            }
        }
        
        if(flag == false) break;
    }
    if(dis1[T] < dis2[T]) printf("WIN\n");
    else if(dis1[T] == dis2[T]) printf("DRAW\n");
    else{
        printf("LOSE\n");
        return 0;
    }
    R(i,1,K){
        printf("%lld ", e[pos[i]].w);
    }
    
    return 0;
}