浙大PAT CCCC L3-014 周游世界 ( 最短路变形 )

题目链接

题意 : 中文题请点链接，挺复杂的...

分析 : 乍一看是个最短路，实际就真的是个最短路。如果没有 “ 在有多条最短路径的时候输出换乘次数最少的” 这一条件的约束，那么这题就是直接建图然后跑个 Dij 就行了，那有了这个约束条件还是要大胆的向最短路思路靠，题目既然需要换乘次数少的，那么我们在进行最短路松弛操作的时候，面对松弛过后最短路径相等的情况就要分开讨论，这时候为了方法取最优值，需要多记录一个信息 ==> 跑到当前点时候换乘次数是多少次，开个数组来记录就行了，其他的还是按最短路来跑。这题就是编码烦了点，不对！是非常烦_(:3 」∠)_

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10;
struct EDGE{ int v, nxt; }Edge[maxn<<2]; ///向前星存图
struct NODE{ ///跑DIJ时塞在优先队列的结构体
    int v, Pre_v, TransferCnt, Dist; ///当前是哪个点、其前一个点是什么、换乘次数、源点到此点的最短距离

    NODE(int V, int D, int Pv, int TCnt):v(V),Dist(D),Pre_v(Pv),TransferCnt(TCnt){};
    bool operator < (const NODE &rhs) const{
        if(this->Dist == rhs.Dist){ ///最短距离相等应当选择换乘次数小的
            return this->TransferCnt > rhs.TransferCnt; ///由于是优先队列、重载小于号需要注意方向....
        }else{
            return this->Dist > rhs.Dist;
        }
    };
};

int cnt; ///边数量
int Head[maxn]; ///邻接表头
int Pre[maxn]; ///答案路径中每个点的前驱、便于恢复路径
int Dis[maxn]; ///记录Dij中源点到其他点的最短路距离
int TransNum[maxn]; ///到达这个点的时候换乘了多少次
int Line[maxn][maxn]; ///记录路线信息
int path[maxn<<2]; ///存储答案路径
bool vis[maxn]; ///DIJ中的标记数组

inline void init() ///初始化表头和计数变量
{
    memset(Head, -1, sizeof(Head));
    cnt = 0;
}

inline void AddEdge(int from, int to) ///加边函数
{
    Edge[cnt].v = to;
    Edge[cnt].nxt = Head[from];
    Head[from] = cnt++;
}

inline void Run_Dijkstra(int st, int en)
{
    memset(vis, false, sizeof(vis));
    memset(Dis, 0x3f3f3f3f, sizeof(Dis));
    memset(TransNum, 0, sizeof(TransNum));
    priority_queue<NODE> que; while(!que.empty()) que.pop();
    Dis[st] = 0;
    que.push(NODE(st,0,0,0));

    while(!que.empty()){
        NODE T = que.top(); que.pop();

        if(vis[T.v]) continue;
        else vis[T.v] = true;

        for(int i=Head[T.v]; i!=-1; i=Edge[i].nxt){
            int Eiv = Edge[i].v;
            if(Dis[Eiv] > Dis[T.v] + 1){ ///满足松弛条件

                Dis[Eiv] = Dis[T.v] + 1;
                int NewTrans = (T.v==st ? 0 : T.TransferCnt + (Line[T.v][Eiv] != Line[T.Pre_v][T.v])); ///计算新的换乘次数

                que.push(NODE(Eiv, Dis[Eiv], T.v, NewTrans));
                Pre[Eiv] = T.v; ///记录前驱、便于恢复路径
                TransNum[Eiv] = NewTrans; ///记录当前点的换乘次数
            }
            else if(Dis[Eiv] == Dis[T.v] + 1 && ///最短距离与松弛后相等则接下来比较换乘次数
                    TransNum[Eiv] > T.TransferCnt + (Line[T.v][Eiv] != Line[T.Pre_v][T.v])){

                que.push(NODE(Eiv, Dis[Eiv], T.v, T.TransferCnt + (Line[T.v][Eiv] != Line[T.Pre_v][T.v])));
                Pre[Eiv] = T.v; ///改变前驱
                TransNum[Eiv] = T.TransferCnt + (Line[T.v][Eiv] != Line[T.Pre_v][T.v]); ///更新换乘次数
            }
        }
    }

    if(Dis[en] == 0x3f3f3f3f){ ///不可达、输出 No Solution
        puts("Sorry, no line is available.");
        return ;
    }else{
        int top = 0; ///记录路径中节点个数
        int now = en; ///由于记录的是前驱、所以从终点开始恢复路径
        int StNext;
        path[top++] = en;
        while(now != st){
            int temp = Pre[now];
            if(temp==st) StNext = now; ///记录起点的后继 ==> 我接下来的输出满足题目所需的答案有需要
            path[top++] = temp;
            now = temp;
        }

        printf("%d\n", Dis[en]); ///先输出最短距离
        int CurLine = Line[st][StNext]; ///从起点开始记录当前所在的铁路编号
        int CurPoint = st; ///当前的点
        for(int i=top-1; i>=1; i--){
            if(Line[path[i]][path[i-1]] == CurLine) continue; ///如果下一个点和仍然在和之前一样的铁路编号则说明不是换乘点
            else{
                printf("Go by the line of company #%d from %04d to %04d.\n",CurLine, CurPoint, path[i]); ///输出格式需要注意....
                ///printf("Go by the line of company #%d from %d to %d.\n", CurLine, CurPoint, path[i]); ///！！！错误的输出格式！！！
                CurPoint = path[i]; ///更新 CurPoint、CurLine
                CurLine = Line[path[i]][path[i-1]];
            }
        }
        printf("Go by the line of company #%d from %04d to %04d.\n",CurLine, CurPoint, path[0]);
       ///printf("Go by the line of company #%d from %d to %d.\n", CurLine, CurPoint, path[0]); ///！！！错误的输出格式！！！
    }
}

int main(void)
{
    init();
    int n;
    scanf("%d", &n);
    for(int i=1; i<=n; i++){
        int num, A, B;
        scanf("%d %d", &num, &A);
        for(int j=1; j<num; j++){
            scanf("%d", &B);
            AddEdge(A, B);
            AddEdge(B, A);
            Line[A][B] = Line[B][A] = i;
            A = B;
        }
    }

    int Query;
    scanf("%d", &Query);
    while(Query--){
        int A, B;
        scanf("%d %d", &A, &B);
        Run_Dijkstra(A, B);
    }
    return 0;
}