PAT 1018 Public Bike Management

不会写，看了别人的思路

先用dijstra保存从PBMC（0结点）到sp节点的最短路径，重点是记录最短路径上的前驱节点

因为最短路径可能不止一条，所以一个节点的前驱节点可能不止一个， 所以要用一个vector来为每个节点维护前驱节点

记录前驱节点后，从sp节点开始向0节点进行dfs遍历，当从sp节点遍历到0节点时，找到了一条最短路径，计算这条路径上的need和back，并更新minneed和minback。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int inf = 99999999;
int e[510][510], weight[510], dis[510];
bool visit[510];
int cmax, n, sp, m;
int minneed = inf, minback = inf;
vector<int> pre[510], path, temppath; //dijstra找到的前驱节点可能不唯一，对每个节点v需要一个vector来记录其前驱节点

void dfs(int v) {
	temppath.push_back(v);
	if (v == 0) {
		int back = 0, need = 0;
		for (int i = temppath.size() - 1; i >= 0; i--) {
			int id = temppath[i];
			if (weight[id] > 0) {
				back += weight[id];
			}
			else {
				if (back > (0 - weight[id])) {
					back += weight[id];
				}
				else {
					need += (0 - weight[id] - back);
					back = 0;
				}
			}
		}
		if (need < minneed) {
			minneed = need;
			minback = back;
			path = temppath;
		}
		else if (need == minneed && back < minback) {
			minback = back;
			path = temppath;
		}
		temppath.pop_back();
		return;
	}
	else {
		for (int i = 0; i < pre[v].size(); i++) {
			dfs(pre[v][i]); //每个节点要循环搜寻其前驱节点，因为它可能不唯一
		}
		temppath.pop_back();
	}

}

int main() {
	fill(e[0], e[0] + 510 * 510, inf);
	fill(dis, dis + 510, inf);
	fill(visit, visit + 510, false);
	cin >> cmax >> n >> sp >> m;
	for (int i = 1; i <= n; i++) {
		cin >> weight[i];
		weight[i] -= cmax / 2;
	}
	for (int i = 0; i < m; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		e[a][b] = e[b][a] = c;
	}
	dis[0] = 0;
	for (int i = 0; i <= n; i++) {
		int u = -1, minn = inf;
		for (int j = 0; j <= n; j++) {
			if (visit[j] == false && dis[j] < minn) {
				u = j;
				minn = dis[j];
			}
		}
		if (u == -1) break;
		visit[u] = true;
		for (int v = 0; v <= n; v++) {
			if (dis[u] + e[u][v] < dis[v]) {
				dis[v] = dis[u] + e[u][v];
				pre[v].clear(); 
				pre[v].push_back(u);
			}
			else if (dis[u] + e[u][v] == dis[v]) {
				pre[v].push_back(u);
			}
		}
	}
	dfs(sp);
	printf("%d 0", minneed);
	for (int i = path.size() - 2; i >= 0; i--) {
		printf("->%d", path[i]);
	}
	printf(" %d", minback);
	return 0;
}

标签：Management,weight,int,back,Bike,1018,510,节点,dis
来源： https://blog.csdn.net/weixin_43480561/article/details/117748193