1018 Public Bike Management （30 分）

There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.

The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in perfect

When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.

The above figure illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S​3​​, we have 2 different shortest paths:

1. PBMC -> S​1​​ -> S​3​​. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S​1​​ and then take 5 bikes to S​3​​, so that both stations will be in perfect conditions.

2. PBMC -> S​2​​ -> S​3​​. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 numbers: C​max​​ (≤100), always an even number, is the maximum capacity of each station; N (≤500), the total number of stations; S​p​​, the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers C​i​​ (i=1,⋯,N) where each C​i​​ is the current number of bikes at S​i​​ respectively. Then M lines follow, each contains 3 numbers: S​i​​, S​j​​, and T​ij​​ which describe the time T​ij​​ taken to move betwen stations S​i​​ and S​j​​. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0−>S​1​​−>⋯−>S​p​​. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of S​p​​ is adjusted to perfect.

Note that if such a path is not unique, output the one that requires minimum number of bikes that we must take back to PBMC. The judge's data guarantee that such a path is unique.

Sample Input:

10 3 3 5
6 7 0
0 1 1
0 2 1
0 3 3
1 3 1
2 3 1

Sample Output:

3 0->2->3 0

一，主要思想

    這道題的最短路徑套模板即可，但是關於第二指標卻不好篩選，題目中是讓路徑長度相同時需要自行車數目越少越好，若是都不需要自行車，則送回總站的自行車數目越少越好。

    首先我考慮的是算出各個站點自行車數目總量和應該佈局自行車的總量進行加減。但是出現的問題是需要甄別是負數的情況，若是負數時需要比較最大值，若是整數時需要比較最小值，還需要加上flag判定，太麻煩。

void DFS(int v) {
	if (v == 0) {
		int sum_num = tempPath.size();
		int sum_half = sum_num*c_max / 2;
		int sum_total = 0;
		for (int i = tempPath.size() - 1; i >= 0; i--) {
			int id = tempPath[i];
			sum_total += current_num[id];
		}
		int middle_1 = sum_half - sum_total;
		if (flag==1) {
			if (middle_1 < 0) {
				flag = 0;
				middle = middle_1;
				path = tempPath;
			}
			else if (middle_1 >= 0 && middle > middle_1) {
				middle = middle_1;
				path = tempPath;
			}
		}
		else if (flag == 0) {
			if (middle_1<0 && middle_1>enmiddle) {
				enmiddle = middle_1;
				path = tempPath;
			}
		}
		return;
	}
	tempPath.push_back(v);
	for (set<int>::iterator it = pre[v].begin(); it != pre[v].end(); it++) {
		DFS(*it);
	}
	tempPath.pop_back();
}

    為了不再flag判定，我選擇了在遍歷中算出總共需要多少，總共返回多少，然後進行比較加減不會出現負數。

void DFS(int v) {
	if (v == 0) {
		int send = 0, take = 0;
		for (int i = tempPath.size() - 1; i >= 0; i--) {
			int id = tempPath[i];
			if (current_num[id] > c_max/2) {
				take += (current_num[id] - c_max / 2);
			}
			else if (current_num[id] < c_max / 2) {
				send += (c_max / 2 - current_num[id]);
			}
		}
		if (send >= take) {
			send -= take;
			take = 0;
		}
		else {
			take -= send;
			send = 0;
		}
		if (send < min_send) {
			min_send = send;
			min_take = take;
			path = tempPath;
		}
		else if (send == min_send&&take < min_take) {
			min_take = take;
			path = tempPath;
		}
		return;
	}
	tempPath.push_back(v);
	for (set<int>::iterator it = pre[v].begin(); it != pre[v].end(); it++) {
		DFS(*it);
	}
	tempPath.pop_back();
}

但是這種方法的前提是考慮往返，所以一個站點若是不夠，可以在返的時候用其他的站點多出來的補上。但題目是隻有往沒有返，所有不能從全域性考慮 。

void DFS(int v) {
	if (v == 0) {
		int len = tempPath.size(), tsum = 0, trequire = 0;
		for (int i = len - 1; i >= 0; i--) {
			if (current_num[tempPath[i]] >= c_max / 2) {
				tsum += current_num[tempPath[i]] - c_max / 2;
			}
			else {
				if (tsum < c_max / 2 - current_num[tempPath[i]]) {
					trequire += (c_max / 2 - current_num[tempPath[i]] - tsum);
					tsum = 0;
				}
				else {
					tsum -= (c_max / 2 - current_num[tempPath[i]]);
				}
			}
		}
		if (trequire < frequire) {
			frequire = trequire;
			path = tempPath;
			freturn = tsum;
		}
		else if (trequire == frequire&&tsum < freturn) {
			path = tempPath;
			freturn = tsum;
		}
		return;
	}
	tempPath.push_back(v);
	for (set<int>::iterator it = pre[v].begin(); it != pre[v].end(); it++) {
		DFS(*it);
	}
	tempPath.pop_back();
}

二，正確程式碼 ：

#include<cstdio>
#include<algorithm>
#include<math.h>
#include<vector>
#include<set>
#include<queue>
using namespace std;
const int max_n = 510;
const int INF = 1000000000;
int	c_max = 0;
int N = 0;
int frequire = INF;
int freturn = INF;
int current_num[max_n];
int times[max_n];
bool vis[max_n] = { false };
int G[max_n][max_n] = { 0 };
vector<int> path, tempPath;
set<int> pre[max_n];
queue<int> que;
int SPFA(int s) {
	fill(times, times + max_n, INF);
	que.push(s);
	times[s] = 0;
	vis[s] = true;
	times[s] = 0;
	while (!que.empty()) {
		int u = que.front();
		que.pop();
		vis[u] = false;
		for (int j = 0; j <= N; j++) {
			if (G[u][j] != 0) {
				int t = G[u][j];
				if (times[u] + t < times[j]) {
					times[j] = times[u] + t;
					pre[j].clear();
					pre[j].insert(u);
					if (!vis[j]) {
						que.push(j);
						vis[j] = true;
					}
				}
				else if (times[u] + t == times[j]) {
					pre[j].insert(u);
				}
			}
		}
	}
}
void DFS(int v) {
	if (v == 0) {
		int len = tempPath.size(), tsum = 0, trequire = 0;
		for (int i = len - 1; i >= 0; i--) {
			if (current_num[tempPath[i]] >= c_max / 2) {
				tsum += current_num[tempPath[i]] - c_max / 2;
			}
			else {
				if (tsum < c_max / 2 - current_num[tempPath[i]]) {
					trequire += (c_max / 2 - current_num[tempPath[i]] - tsum);
					tsum = 0;
				}
				else {
					tsum -= (c_max / 2 - current_num[tempPath[i]]);
				}
			}
		}
		if (trequire < frequire) {
			frequire = trequire;
			path = tempPath;
			freturn = tsum;
		}
		else if (trequire == frequire&&tsum < freturn) {
			path = tempPath;
			freturn = tsum;
		}
		return;
	}
	tempPath.push_back(v);
	for (set<int>::iterator it = pre[v].begin(); it != pre[v].end(); it++) {
		DFS(*it);
	}
	tempPath.pop_back();
}
int main() {
	int broken = 0, M = 0;
	int x = 0, y = 0, z = 0;
	scanf("%d %d %d %d", &c_max, &N, &broken, &M);
	for (int i = 1; i <= N; i++) {
		scanf("%d", &current_num[i]);
	}
	for (int i = 0; i < M; i++) {
		scanf("%d %d %d", &x, &y, &z);
		G[x][y] = z;
		G[y][x] = z;
	}
	SPFA(0);
	DFS(broken);
	printf("%d ", frequire);
	printf("0->");
	for (int i = path.size() - 1; i >= 0; i--) {
		printf("%d", path[i]);
		if (i != 0) {
			printf("->");
		}
		else {
			printf(" ");
		}
	}
	printf("%d", freturn);
	return 0;
}