1018 Public Bike Management (30point(s)) 需要二刷 *DFS+DIJ多路径保存问题

#include<iostream>
#include<fstream>
#include<sstream>
#include<vector>
#include<string>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<string.h>
using namespace std;

const int maxn = 510;
const int INF = 100000;

int cmax, n, sp, p, m;
int sta[maxn];
int ma[maxn][maxn];
bool vis[maxn];
int dis[maxn];
vector<int> pre[maxn];
vector<int> tempPath;//临时路径；
vector<int> path;//最终保留路径；
int numpath;
int minNeed = INF;
int minRemain = INF;

void dij(int st) {
    fill(vis, vis + maxn, true);
    fill(dis, dis + maxn, INF);
    dis[st] = 0;
    for (int i = 0; i <= n; i++) {
        int index = -1;
        int min = INF;
        for (int j = 0; j <= n; j++) {
            //寻找最小节点；
            if (vis[j]&&min > dis[j]) {
                min = dis[j];
                index = j;
            }
        }
        if (index == -1)
            return;
        //进行后续的寻找；
        vis[index] = false;
        //cout << 1 << endl;
        for (int j = 0; j <= n; j++) {
            if (vis[j] && ma[index][j] != INF) {
                if (dis[j] > dis[index] + ma[index][j]) {
                    //如果存在更小路径;
                    dis[j] = dis[index] + ma[index][j];
                    pre[j].clear();
                    pre[j].push_back(index);
                }
                else if (dis[j] == dis[index] + ma[index][j]) {
                    //如果存在相等路径；
                    pre[j].push_back(index);
                }
            }
        }
    }
}

void DFS(int v) {
    if (v == 0) {
        //v为开始节点， 代表已经到了尽头；
        tempPath.push_back(v);
        int need = 0;
        int remain = 0;
        for (int i = tempPath.size()-1; i >= 0; i--) {
            //相当于从根开始遍历节点；
            int id = tempPath[i];
            if (sta[id] > 0) {
                //需要带走自行车；
                remain += sta[id];
            }
            else {
                if (remain > abs(sta[id])) {
                    //如果持有量大于补给；
                    remain -= abs(sta[id]);
                }
                else {
                    //需要补充
                    need += abs(sta[id]) - remain;
                    remain = 0;
                }
            }
        }
        if (need < minNeed) {
            minNeed = need;
            minRemain = remain;
            path = tempPath;
        }
        else if (need == minNeed && remain < minRemain) {
            minRemain = remain;
            path = tempPath;
        }
        tempPath.pop_back();
        return;
    }
    tempPath.push_back(v);
    for (int i = 0; i < pre[v].size(); i++) {
        DFS(pre[v][i]);
    }
    tempPath.pop_back();
}

int main(){
    int a, b;
    fill(ma[0], ma[0] + maxn * maxn, INF);
    cin >> cmax >> n >> sp >> m;
    sta[0] = 0;
    for (int i = 1; i <=n ; i++) {
        cin >> sta[i];
        sta[i] -= cmax / 2;
    }
    for (int i = 0; i < m; i++) {
        cin >> a >> b;
        cin >> ma[a][b];
        ma[b][a] = ma[a][b];
    }
    dij(0);
    DFS(sp);
    printf("%d ", minNeed);
    for (int i = path.size() - 1; i >= 0; i--) {
        printf("%d", path[i]);
        if (i > 0)
            printf("->");
    }
    printf(" %d", minRemain);
    return 0;
}