CF360E Levko and Game（贪心）

这题贪心停水的，找\(dis1<=dis2\)的点往歇斯底里地砍，砍到没法砍就是。

写博客是为了记录下遇到的神奇bug

#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;
}

cdecl：

sizeof 一个指针肯定会死啊

这样只会 memset 前 4（64 位机的话，8）个 bit