洛谷 P1073

题目链接:P1073 最优贸易

题目大意

在一号城市开始走,一直走到n号城市,可以在路径上买东西和卖东西,使其赚的钱最大, 也可以不买不卖

solution

第一眼题目 (Tarjan) 缩点?写起来是不是有点长啊

那我们呢就用分层图做吧,然后跑一个 (SPFA) (不是死了么/微笑)

每一层的连边都为 (0) ,然后第一层想第二层连边代表买入为负的,第二层向第三层连边代表卖出,然后在与超级终点连边,总此三层,然后跑最长路

code:

/**
*    Author: Alieme
*    Data: 2020.8.30
*    Problem: P1073
*    Time: O()
*/
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>

#define ll long long
#define rr register

#define inf 1e9
#define MAXN 100010

using namespace std;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void print(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + 48);
}

struct Node {
	int v;
	int len;
	Node() {}
	Node(int V, int Len) {v = V, len = Len;}
};

int n, m;

int v[MAXN], d[MAXN << 2];

vector<Node> G[MAXN << 2];

void add(int x, int y) {
	G[x].push_back(Node(y, 0));
	G[x + n].push_back(Node(y + n, 0));
	G[x + 2 * n].push_back(Node(y + 2 * n, 0));
	G[x].push_back(Node(y + n, -v[x]));
	G[x + n].push_back(Node(y + 2 * n, v[x]));
}

queue<int> q;

bool inq[MAXN << 2];

void SPFA() {
	for (rr int i = 1; i <= n; i++) d[i] = -inf;
	d[1] = 0;
	inq[1] = true;
	q.push(1);
	while (!q.empty()) {
		int tp = q.front();
		q.pop();
		inq[tp] = false;
		int len = G[tp].size();
		for (rr int i = 0; i < len; i++) {
			Node x = G[tp][i];
			if (d[x.v] < d[tp] + x.len) {
				d[x.v] = d[tp] + x.len;
				if (inq[x.v] == false) {
					q.push(x.v);
					inq[x.v] = true;
				}
			}
		}
	}
}

signed main() {
	n = read();
	m = read();
	for (rr int i = 1; i <= n; i++) v[i] = read();
	for (rr int i = 1; i <= m; i++) {
		int x = read();
		int y = read();
		add(x, y);
		int z = read();
		if (z == 2) add(y, x);
	}
	G[n].push_back(Node(3 * n + 1, 0));
	G[n * 3].push_back(Node(3 * n + 1, 0));
	n = 3 * n + 1;
	SPFA();
	cout << d[n];
}