2021-03-15

北大 算法基础
深搜之寻路

练习代码:

#include<iostream> 
#include<vector>
using namespace std;

int K, N, R, S, D, L, T;//开始城市,到达城市,路长,花费
struct Road {
	int d, l, t;
};
vector<vector<Road>> cityMaps(110);//城市邻接表,内vector保存各城市可到达的城市,路长,花费
int minLen = 1 << 30;//到终点的最短路长
int totalLen, totalCost;//到当前节点的路长和花费
int visited[110];//标记已走过的城市
int minL[110][10100];//走到各节点同样花费的最短路长

void Dfs(int s) {
	if (s == N) {//到达终点
		minLen = min(minLen, totalLen);
		return;
	}
	for (int i = 0; i < cityMaps[s].size(); i++) {
		int d = cityMaps[s][i].d;//下一个城市
		if (!visited[d]) {
			int cost = totalCost + cityMaps[s][i].t;//如果走此节点的总花费
			if (cost > K) continue;//超支,放弃
			if (totalLen + cityMaps[s][i].l >= minLen || totalLen + cityMaps[s][i].l >= minL[d][cost])
				continue;//超过最优路长,超过到d的cost花费的路长,放弃
			//符合条件走到d
			totalLen += cityMaps[s][i].l;
			totalCost += cityMaps[s][i].t;
			minL[d][cost] = totalLen;
			visited[d] = 1;
			Dfs(d);
			//恢复走到d前的状态
			visited[d] = 0;
			totalCost -= cityMaps[s][i].t;
			totalLen -= cityMaps[s][i].l;
		}
	}
}

int main() {
	cout << "总金额,总城市数,总路数:";
	cin >> K >> N >> R;//总金额,总城市数,总路数
	cout << endl << "起点数值,到达点,路长,花费:" << endl;
	for (int i = 0; i < R; i++)
	{
		int s;//起点
		Road r;
		cout << i+1 << "): ";
		cin >> s >> r.d >> r.l >> r.t;
		if (s != r.d)
			cityMaps[s].push_back(r);//s城市节点的所有路信息
	}
	for (int i = 0; i < 110; i++)
		for (int j = 0; j < 10100; j++)
			minL[i][j] = 1 << 30;
	memset(visited, 0, sizeof(visited));
	totalLen = 0;
	totalCost = 0;
	visited[1] = 1;
	minLen = 1 << 30;//最优路长赋极大值
	Dfs(1);
	if (minLen < (1 << 30))
		cout << minLen << endl;
	else
		cout << "-1" << endl;

	return 0;
}

输入输出样例:

标签：03,15,minLen,int,路长,totalLen,cityMaps,2021,cost
来源： https://blog.csdn.net/ywqzj/article/details/114852847