BZOJ2118 墨墨的等式 【最短路】

题目链接

BZOJ2118

题解

orz竟然是最短路

我们去(0)后取出最小的(a[i])，记为(p)，然后考虑模(p)下的(B)
一个数(i)能被凑出，那么(i + p)也能被凑出
所以我们只需找出最小的凑出(i)的代价

我们如果将同余下的和看作点，那么加上一个数就相当于在点间转移的边
所以我们只需跑最短路即可求出每个(i)的最小代价，然后就可以计算(Bmin)和(Bmax)以内分别有多少个(i)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 500005,maxm = 5000005;
const LL INF = 100000000000000001ll;
inline LL read(){
	LL out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
struct node{
	int u; LL d;
};
inline bool operator <(const node& a,const node& b){
	return a.d > b.d;
}
inline bool operator ==(const node& a,const node& b){
	return a.u == b.u && a.d == b.d;
}
struct Heap{
	priority_queue<node> a,b;
	void ck(){while (!b.empty() && a.top() == b.top()) a.pop(),b.pop();}
	int size(){return a.size() - b.size();}
	node top(){ck(); node x = a.top(); a.pop(); return x;}
	void del(node x){ck(); b.push(x);}
	void ins(node x){ck(); a.push(x);}
}H;
int N,a[maxn],P;
LL d[maxn]; int vis[maxn];
int h[maxn],ne;
struct EDGE{int to,nxt,w;}ed[maxm];
inline void build(int u,int v,int w){
	ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;
}
void work(){
	for (int i = 0; i < P; i++){
		for (int j = 1; j <= N; j++)
			build(i,(i + a[j]) % P,a[j]);
	}
	for (int i = 1; i < P; i++) d[i] = INF;
	d[0] = 0; H.ins((node){0,d[0]}); vis[0] = true;
	node u;
	while (H.size()){
		u = H.top();
		Redge(u.u) if (!vis[to = ed[k].to] && d[to] > d[u.u] + ed[k].w){
			if (d[to] != INF) H.del((node){to,d[to]});
			d[to] = d[u.u] + ed[k].w;
			H.ins((node){to,d[to]});
		}
	}
}
int main(){
	N = read(); LL L = read(),R = read(); P = INF;
	REP(i,N){
		a[i] = read();
		if (!a[i]) i--,N--;
	}
	if (!N){
		if (L) puts("0");
		else puts("1");
		return 0;
	}
	REP(i,N) P = min(P,a[i]);
	work();
	L--;
	LL ansl = 0,ansr = 0;
	for (int i = 0; i < P; i++){
		if (d[i] <= L){
			ansl++;
			ansl += (L - d[i]) / P;
		}
		if (d[i] <= R){
			ansr++;
			ansr += (R - d[i]) / P;
		}
	}
	printf("%lld
",ansr - ansl);
	return 0;
}