算法训练营|图论第10天 Bellman_ford:优化算法，判断负权算法，单源有限最短路

题目：Bellman_ford:优化算法

题目链接：

94. 城市间货物运输 I (kamacoder.com)

代码：

#include<bits/stdc++.h>
using namespace std;
struct Edge {
	int to;
	int val;
	Edge(int t, int w) :to(t), val(w) {}
};
int main() {
	int n, m;
	cin >> n >> m;
	int p1, p2, val;
	vector<list<Edge>>grid(n + 1);
	for (int i = 0; i < m; i++) {
		cin >> p1 >> p2 >> val;
		grid[p1].push_back(Edge(p2, val));
	}
	int start = 1;
	int end = n;

	queue<int>que;
	vector<bool>InQueue(n + 1, false);
	vector<int>minDist(n + 1, INT_MAX);
	minDist[start] = 0;
	que.push(start);
	while (!que.empty()) {
		int cur = que.front();
		que.pop();
		InQueue[cur] = false;
		for (Edge edge : grid[cur]) {
			int from = cur;
			int to = edge.to;
			int val = edge.val;
			if (minDist[to] > minDist[from] + val) {
				minDist[to] = minDist[from] + val;
				if (InQueue[to] == false) {
					que.push(to);
					InQueue[to] = true;
				}
			}
		}
	}
	if (minDist[end] == INT_MAX) cout << "unconnected" << endl;
	else cout << minDist[end] << endl;
}

题目：判断负权算法

题目链接：

95. 城市间货物运输 II (kamacoder.com)

代码：

#include<bits/stdc++.h>
using namespace std;
struct Edge {
	int to;
	int val;
	Edge(int t, int w) :to(t), val(w) {}
};
int main() {
	int n, m;
	cin >> n >> m;
	int p1, p2, val;
	vector<vector<int>>grid;
	for (int i = 0; i < m; i++) {
		cin >> p1 >> p2 >> val;
		grid.push_back({ p1,p2,val });
	}
	int start = 1;
	int end = n;

	vector<int>minDist(n + 1, INT_MAX);
	minDist[start] = 0;
	bool flag = false;
	for (int i = 1; i <= n ; i++) {
		for (vector<int>vec : grid) {
			int from = vec[0];
			int to = vec[1];
			int val = vec[2];
			if (i < n) {
				if (minDist[from] != INT_MAX && minDist[from] + val < minDist[to]) {
					minDist[to] = minDist[from] + val;
				}
			}
			else {
				if (minDist[from] != INT_MAX && minDist[from] + val < minDist[to]) {
					flag = true;
				}
			}
			
		}
	}
	if (flag) {
		cout << "circle" << endl;
	}
	else if (minDist[end] == INT_MAX) {
		cout << "unconnected" << endl;
	}
	 else {
		 cout << minDist[end] << endl;
	 }
}

题目：单源有限最短路

题目链接：

96. 城市间货物运输 III (kamacoder.com)

代码：

#include<bits/stdc++.h>
using namespace std;
struct Edge {
	int to;
	int val;
	Edge(int t, int w) :to(t), val(w) {}
};
int main() {
	int n, m;
	cin >> n >> m;
	int p1, p2, val;
	vector<vector<int>>grid;
	for (int i = 0; i < m; i++) {
		cin >> p1 >> p2 >> val;
		grid.push_back({ p1,p2,val });
	}
    int src, dst, k;
    cin >> src >> dst >> k;

    vector<int> minDist(n + 1, INT_MAX);
    minDist[src] = 0;
    vector<int> minDist_copy(n + 1); // 用来记录上一次遍历的结果
	for (int i = 1; i <= k + 1; i++) {
		minDist_copy = minDist;
		for (vector<int> &vec : grid) {
			int from = vec[0];
			int to = vec[1];
			int val = vec[2];
			if (minDist_copy[from] != INT_MAX && minDist[to] > minDist_copy[from] + val) {
				minDist[to] = minDist_copy[from] + val;
			}
		}
	}
    if (minDist[dst] == INT_MAX) cout << "unreachable" << endl; // 不能到达终点
    else cout << minDist[dst] << endl; // 到达终点最短路径
}