[NOI2020]美食家 题解

考虑拆点(,)把每个点拆成(5)个点(,) (p_{i,t}(0leq t leq 4))表示(t)秒之后到达点(i.)

对于每个正整数(t,)把所有的(p_{i,t})连向(p_{i,t-1}.)

然后每一条边((x,y,t))就可以变成((p_{x,0},p_{y,t-1}),)就可以矩阵乘法来维护这个东西了(.)

不难据此得到转移矩阵(A.)

然后我们相当于要向量乘矩阵(,)一共需要乘(k)次(A)的若干次方(,)那么我们预处理一下(A^2,A^4,A^8,...A^{2^{30}})即可(.)

复杂度(O((nw)^3 log T+k(nw)^2 log T).)

代码(:)

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 50,M = 505,K = 205,V = 250;
const LL INF = 1ll << 60;

int L;
struct Mat{
	LL a[V][V];
	inline void init(){
		for (int i = 0; i < L; ++i) for (int j = 0; j < L; ++j) a[i][j] = -INF;
	}
};
struct Vec{
	LL a[V];
	inline void init(){
		for (int i = 0; i < L; ++i) a[i] = -INF;	
	}
};
Mat operator * (const Mat A,const Mat B){
	static Mat T; T.init();
	for (int k = 0; k < L; ++k) for (int i = 0; i < L; ++i) for (int j = 0; j < L; ++j)
		if (A.a[i][k] + B.a[k][j] > T.a[i][j]) T.a[i][j] = A.a[i][k] + B.a[k][j];
	return T;
}
Vec operator * (const Mat A,const Vec B){
	static Vec T; T.init();
	for (int i = 0; i < L; ++i) for (int j = 0; j < L; ++j)
		if (A.a[j][i] + B.a[j] > T.a[i]) T.a[i] = A.a[j][i] + B.a[j];
	return T;
}

int n,m,T,k,c[N];
int ex[M],ey[M],ez[M];
Mat I,Tr,A[30];
Vec st; int nowt;
int id[N][5];
struct Festival{
	int t,x,y;
}ev[K];
int main(){
//	freopen("delicacy.in","r",stdin);
//	freopen("delicacy.out","w",stdout);
	int i,j;	
	cin >> n >> m >> T >> k;
	I.init(); L = n * 5;
	for (i = 0; i < n; ++i) cin >> c[i];
	for (i = 0; i < n; ++i) I.a[i][i] = 0;
	for (i = 0; i < n; ++i) for (j = 0; j < 5; ++j) id[i][j] = j * n + i;
	for (i = 1; i <= m; ++i) cin >> ex[i] >> ey[i] >> ez[i],--ex[i],--ey[i];
	Tr.init();
	for (i = 0; i < n; ++i)
	for (j = 1; j < 5; ++j) Tr.a[id[i][j]][id[i][j-1]] = (j==1) ? (c[i]) : (0);
	for (i = 1; i <= m; ++i) Tr.a[id[ex[i]][0]][id[ey[i]][ez[i]-1]] = (ez[i]==1) ? (c[ey[i]]) : (0);
	A[0] = Tr;
	for (i = 1; i < 30; ++i) A[i] = A[i-1] * A[i-1];
	
	for (i = 1; i <= k; ++i) cin >> ev[i].t >> ev[i].x >> ev[i].y,--ev[i].x;
	for (i = 1; i <= k; ++i) for (j = i+1; j <= k; ++j) if (ev[j].t < ev[i].t) swap(ev[i],ev[j]);
	if (ev[k].t != T){ ++k; ev[k].t = T,ev[k].x = ev[k].y = 0; }	
	st.init(); st.a[id[0][0]] = c[0]; nowt = 0;
	for (i = 1; i <= k; ++i){
		int dt = ev[i].t - nowt;
		for (j = 0; j < 30; ++j) if (dt>>j&1) st = A[j] * st;
		if (st.a[ev[i].x] >= 0) st.a[ev[i].x] += ev[i].y; 
		nowt = ev[i].t;
	}
	if (st.a[0] < 0) st.a[0] = -1;
	cout << st.a[0] << '
';
}