Codeforces 576D Flights for Regular Customers (图论、矩阵乘法、Bitset)

题目链接

http://codeforces.com/contest/576/problem/D

题解

把边按(t_i)从小到大排序后枚举(i), 求出按前((i-1))条边走(t_i)步能到达的点的集合，以它们为起点求(n)号点的最短路。
前者等于前((i-2))条边走(t_{i-1})步能到达的点集乘上前((i-1))条边邻接矩阵的((t_i-t_{i-1}))次幂。
因为只关心是否存在，故可以使用bitset优化。
时间复杂度(O(mn^3+frac{n^3log T}{omega})).

代码

#include<bits/stdc++.h>
using namespace std;

inline int read()
{
	int x = 0,f = 1; char ch = getchar();
	for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}
	for(;!isdigit(ch);ch=getchar()) {x = x*10+ch-48;}
	return x*f;
}

const int N = 150;
const int INF = 2e9;
struct AEdge
{
	int u,v,t;
	bool operator <(const AEdge &arg) const {return t<arg.t;}
} ae[N+3];
struct Edge
{
	int v,nxt;
} e[(N<<1)+3];
int a[N+3];
int fe[N+3];
int que[N+3];
int dep[N+3];
bool vis[N+3];
int n,m,en;

void addedge(int u,int v)
{
	en++; e[en].v = v;
	e[en].nxt = fe[u]; fe[u] = en;
}

struct Matrix
{
	bitset<N+3> a[N+3];
	Matrix() {for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) a[i][j] = 0;}
	Matrix operator *(const Matrix &arg) const
	{
		Matrix ret;
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=n; j++)
			{
				if(a[i][j]) {ret.a[i]|=arg.a[j];}
			}
		}
		return ret;
	}
};
Matrix I,O,g,b;

void clearedge()
{
	for(int i=1; i<=n; i++) fe[i] = 0;
	for(int i=1; i<=en; i++) e[i].v = e[i].nxt = 0;
	en = 0;
}

Matrix mquickpow(Matrix x,int y)
{
	Matrix cur = x,ret = I;
	for(int i=0; y; i++)
	{
		if(y&(1<<i)) {y-=(1<<i); ret = ret*cur;}
		cur = cur*cur;
	}
	return ret;
}

void bfs()
{
	for(int i=1; i<=n; i++) vis[i] = false;
	int hd = 1,tl = 0;
	for(int i=1; i<=n; i++) {if(g.a[1][i]) {tl++; que[tl] = i; vis[i] = true; dep[i] = 0;}}
	while(hd<=tl)
	{
		int u = que[hd]; hd++;
		for(int i=fe[u]; i; i=e[i].nxt)
		{
			int v = e[i].v;
			if(!vis[v]) {vis[v] = true; tl++; que[tl] = v; dep[v] = dep[u]+1;}
		}
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1; i<=n; i++) I.a[i].set(i);
	for(int i=1; i<=m; i++)
	{
		scanf("%d%d%d",&ae[i].u,&ae[i].v,&ae[i].t);
	}
	sort(ae+1,ae+m+1);
	g = I;
	int ans = INF;
	for(int i=1; i<=m; i++)
	{
		for(int j=1; j<=i; j++) {addedge(ae[j].u,ae[j].v);}
		g = g*mquickpow(b,ae[i].t-ae[i-1].t);
		bfs();
		if(vis[n]) {ans = min(ans,ae[i].t+dep[n]);}
		clearedge();
		b.a[ae[i].u].set(ae[i].v);
	}
	if(ans==INF) puts("Impossible");
	else printf("%d
",ans);
	return 0;
}