BZOJ 4386 Luogu P3597 [POI2015]Wycieczki (矩阵乘法)

题目链接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=4386
(luogu) https://www.luogu.org/problemnew/show/P3597

为啥这种题我都不会了啊
题解: 首先如果边权全都为\(1\), 那么就新建一个计数器，每个点连计数器，计数器连个自环。然后邻接矩阵快速幂倍增即可
如果边权有\(2\)和\(3\), 就分别新建一个节点连向出点
细节不少，特别是判断是否大于\(k\)的时候不能爆long long(据说这题数据水，所以我不敢保证下面的代码不会被卡)

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iostream>
#define llong long long
using namespace std;

inline int read()
{
	int x=0; bool f=1; char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=0;
	for(; isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
	if(f) return x;
	return -x;
}

const int N = 121;
llong p;
struct Matrix
{
	llong a[N+3][N+3]; int n;
	Matrix() {}
	Matrix(int _n) {n = _n; for(int i=0; i<=n; i++) for(int j=0; j<=n; j++) a[i][j] = 0ll;}
	void init(int _n) {n = _n; for(int i=0; i<=n; i++) for(int j=0; j<=n; j++) a[i][j] = 0ll;}
	void unitize(int _n) {n = _n; a[0][0] = 0ll; for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) a[i][j] = i==j?1ll:0ll;}
	void output() {if(a[0][0]==-1) puts("gg"); for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) printf("%d ",a[i][j]); puts("");}}
	Matrix operator *(const Matrix &arg) const
	{
		Matrix ret(n);
		for(int i=1; i<=n; i++)
		{
			for(int j=1; j<=n; j++)
			{
				for(int k=1; k<=n; k++)
				{
					if(k==n)
					{
						if(arg.a[j][k]>0 && (a[i][j]>=p/arg.a[j][k]+1 || ret.a[i][k]>=p-a[i][j]*arg.a[j][k]))
						{
							ret.a[0][0] = -1;
							return ret;
						}
					}
					ret.a[i][k] = ret.a[i][k]+a[i][j]*arg.a[j][k];
				}
			}
		}
		llong sum = 0ll;
		for(int i=1; i<=n/3; i++)
		{
			if(sum>=p-ret.a[i][n]) {ret.a[0][0] = -1; return ret;}
			sum += ret.a[i][n];
		}
		return ret;
	}
} g,pw[65],tmp;
int n,m;

llong solve()
{
	pw[0] = g; int i = 0;
	for(i=1; i<=61; i++)
	{
		pw[i] = pw[i-1]*pw[i-1];
		if(pw[i].a[0][0]==-1)
		{
			break;
		}
	}
	if(i==62) {return -1;}
	llong ret = 0ll; g.unitize(n+n+n+1);
	for(i--; i>=0; i--)
	{
		tmp = g*pw[i];
		if(tmp.a[0][0]!=-1)
		{
			ret|=(1ll<<i);
			g = tmp;
		}
	}
	return ret;
}

int main()
{
	scanf("%d%d%lld",&n,&m,&p); p+=n;
	g.init(n+n+n+1);
	for(int i=1; i<=m; i++)
	{
		int x,y,w; scanf("%d%d%d",&x,&y,&w);
		if(w==1) {g.a[x][y]++;}
		else if(w==2) {g.a[x+n][y]++;}
		else if(w==3) {g.a[x+n+n][y]++;}
	}
	for(int i=1; i<=n; i++)
	{
		g.a[i][i+n]++;
		g.a[i+n][i+n+n]++;
		g.a[i][n+n+n+1]++;
	}
	g.a[n+n+n+1][n+n+n+1]++;
	llong ans = solve();
	printf("%lld\n",ans);
	return 0;
}