题目描述

https://loj.ac/problem/539

题解

显然是跑到没油了才加油，所以设f[i,j]表示从点i开始钱为j的最大距离，逆着做

转移考虑设g[i,j,k]表示从i到j走2^k步的答案，倍增求

再求出w[i,j,k]表示从i到j走k步的答案，拆位后做log次

最后二分答案

code

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define ll long long
#define file
using namespace std;

int P[17],a[1001][3],ls[101],p[101],c[101],h,t,n,m,C,Q,i,j,k,l,r,mid,len,x,y,z,s;
ll g[101][101][17],w[101][101][17],f[10001][101],ans;

void New(int x,int y,int z) {++len;a[len][0]=y;a[len][1]=ls[x];ls[x]=len;a[len][2]=z;}
void work1()
{
	fo(l,1,16)
	{
		fo(i,1,n) fo(j,1,n) g[i][j][l]=g[i][j][l-1];
		fo(i,1,n)
		{
			fo(j,1,n)
			{
				fo(k,1,n)
				g[i][j][l]=max(g[i][j][l],g[i][k][l-1]+g[k][j][l-1]);
			}
		}
	}
}
void work2()
{
	fo(i,1,n)
	{
		w[i][i][0]=0;
		s=min(C,c[i]);
		if (s&1)
		{
			fo(j,1,n)
			w[i][j][0]=g[i][j][0];
		}
	}
	fo(l,1,16)
	{
		fo(i,1,n)
		{
			s=min(C,c[i]);
			fo(j,1,n)
			{
				w[i][j][l]=w[i][j][l-1];
				if (s&P[l])
				{
					fo(k,1,n)
					w[i][j][l]=max(w[i][j][l],w[i][k][l-1]+g[k][j][l]);
				}
			}
		}
	}
}
void work3()
{
	fo(i,1,n) f[0][i]=0;
	fo(i,1,n*n)
	{
		fo(j,1,n)
		{
			f[i][j]=f[i-1][j];
			if (i>=p[j])
			{
				fo(k,1,n)
				f[i][j]=max(f[i][j],w[j][k][16]+f[i-p[j]][k]);
			}
		}
	}
}

int main()
{
	freopen("trip.in","r",stdin);
	#ifdef file
	freopen("trip.out","w",stdout);
	#endif
	
	memset(f,200,sizeof(f));
	memset(g,200,sizeof(g));
	memset(w,200,sizeof(w));
	P[0]=1;
	fo(i,1,16) P[i]=P[i-1]*2;
	scanf("%d%d%d%d",&n,&m,&C,&Q);
	fo(i,1,n) scanf("%d%d",&p[i],&c[i]),g[i][i][0]=0;
	fo(i,1,m) scanf("%d%d%d",&j,&k,&l),New(j,k,l),g[j][k][0]=max(g[j][k][0],l);
	
	work1();
	work2();
	work3();
	
	for (;Q;--Q)
	{
		scanf("%d%d%d",&x,&y,&z);
		l=0,r=y;
		while (l<r)
		{
			mid=(l+r)/2;
			if (f[mid][x]<z)
			l=mid+1; else r=mid;
		}
		printf("%d\n",(f[l][x]<z)?-1:y-l);
	}
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}