PAT1018. Public Bike Management (30)

Posted 2020-09-27 肉松松鼠

题目传送门

分析

　　Dijstra + DFS搜索，难点在于存在多条最短路径，因此需要用vector<int> pre[N] 记录路径。

代码

　　自己写的有些问题，有两个数据没过，就不放上来了。下面的代码来自于https://www.liuchuo.net/archives/2373

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 99999999;
int cmax, n, sp, m;
int minNeed = inf, minBack = inf;
int e[510][510], dis[510], weight[510];
bool visit[510];
vector<int> pre[510];
vector<int> path, temppath;
void dfs(int v) {
    if(v == 0) {
        temppath.push_back(v);
        int need = 0, back = 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 ;
    }
    temppath.push_back(v);
    for(int i = 0; i < pre[v].size(); i++)
        dfs(pre[v][i]);
    temppath.pop_back();
}
int main() {
    fill(e[0], e[0] + 510 * 510, inf);
    fill(dis, dis + 510, inf);
    scanf("%d%d%d%d", &cmax, &n, &sp, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &weight[i]);
        weight[i] = weight[i] - cmax / 2;
    }
    for(int i = 0; i < m; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        scanf("%d", &e[a][b]);
        e[b][a] = e[a][b];
    }
    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(visit[v] == false && e[u][v] != inf) {
                if(dis[v] > dis[u] + e[u][v]) {
                    dis[v] = dis[u] + e[u][v];
                    pre[v].clear();　　//这里要记得clear
                    pre[v].push_back(u);
                }else if(dis[v] == dis[u] + e[u][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;
}