L3-007. 天梯地图

时间限制

300 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

做者

陈越

本题要求你实现一个天梯赛专属在线地图，队员输入本身学校所在地和赛场地点后，该地图应该推荐两条路线：一条是最快到达路线；一条是最短距离的路线。题目保证对任意的查询请求，地图上都至少存在一条可达路线。ios

输入格式：spa

输入在第一行给出两个正整数N（2 <= N <=500）和M，分别为地图中全部标记地点的个数和链接地点的道路条数。随后M行，每行按以下格式给出一条道路的信息：code

V1 V2 one-way length timeblog

其中V1和V2是道路的两个端点的编号（从0到N-1）；若是该道路是从V1到V2的单行线，则one-way为1，不然为0；length是道路的长度；time是经过该路所须要的时间。最后给出一对起点和终点的编号。ip

输出格式：内存

首先按下列格式输出最快到达的时间T和用节点编号表示的路线：string

Time = T: 起点 => 节点₁ => ... => 终点it

而后在下一行按下列格式输出最短距离D和用节点编号表示的路线：io

Distance = D: 起点 => 节点₁ => ... => 终点class

若是最快到达路线不惟一，则输出几条最快路线中最短的那条，题目保证这条路线是惟一的。而若是最短距离的路线不惟一，则输出途径节点数最少的那条，题目保证这条路线是惟一的。

若是这两条路线是彻底同样的，则按下列格式输出：

Time = T; Distance = D: 起点 => 节点₁ => ... => 终点

输入样例1：

输出样例1：

Time = 6: 5 => 4 => 8 => 3
Distance = 3: 5 => 1 => 3

输入样例2:

输出样例2:

Time = 3; Distance = 4: 3 => 2 => 5

思路：最短路，不过麻烦的是多条最短路中推荐最优的路线，并输出路径。路径的还原能够不断记录前驱节点，注意的是每一个节点的前驱节点可能不止一个，全须要记录，最后dfs搜索最优路径。

#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<vector> #include<string> #include<iomanip> #include<map> #include<stack> #include<set> #include<queue>
using namespace std; #define N_MAX 500+20
#define INF 0x3f3f3f3f
int n, m; int s, t; struct edge { int to, cost_t, cost_l; edge() {} edge(int to,int cost_t,int cost_l):to(to),cost_t(cost_t),cost_l(cost_l) {} }; struct P { int first, second;//first距离，second节点编号
 P() {} P(int first,int second):first(first),second(second) {} bool operator < (const P&b) const{ return first > b.first; } }; vector<edge>G[N_MAX]; int d_t[N_MAX], d_l[N_MAX]; vector<int>prev_l[N_MAX];//记录最短路径的前驱结点，每一个点均可能会有几个前驱结点
vector<int>prev_t[N_MAX];//记录最短时限路径的前驱结点，同上
int Dist[N_MAX][N_MAX];//邻接矩阵

void dijkstra1(int s) { priority_queue<P>que; fill(d_l,d_l+n,INF); d_l[s] = 0; que.push(P(0, s)); while (!que.empty()) { P p = que.top(); que.pop(); int v = p.second; if (d_l[v] < p.first)continue; for (int i = 0; i < G[v].size();i++) { edge e = G[v][i]; if (d_l[e.to] > d_l[v] + e.cost_l) { d_l[e.to] = d_l[v] + e.cost_l; que.push(P(d_l[e.to], e.to)); prev_l[e.to].clear(); prev_l[e.to].push_back(v); } else if (d_l[e.to] == d_l[v] + e.cost_l) { prev_l[e.to].push_back(v); } } } } void dijkstra2(int s) { priority_queue<P>que; fill(d_t, d_t + n, INF); d_t[s] = 0; que.push(P(0, s)); while (!que.empty()) { P p = que.top(); que.pop(); int v = p.second; if (d_t[v] < p.first)continue; for (int i = 0; i < G[v].size(); i++) { edge e = G[v][i]; if (d_t[e.to] > d_t[v] + e.cost_t) { d_t[e.to] = d_t[v] + e.cost_t; que.push(P(d_t[e.to], e.to)); prev_t[e.to].clear(); prev_t[e.to].push_back(v); } else if (d_t[e.to] == d_t[v] + e.cost_t) { prev_t[e.to].push_back(v); } } } } int road2[N_MAX]; vector<int>r2; int min_step = INF; void dfs2(int x,int step) {//最短距离同样，取节点最少的路径
    road2[step] = x; if (x == s) {//到达起点
        if (min_step > step) { min_step = step; r2.clear(); for (int i = step; i >= 0; i--)r2.push_back(road2[i]); } return; } for (int i = 0; i < prev_l[x].size();i++) { dfs2(prev_l[x][i], step + 1); } } int road1[N_MAX]; vector<int>r1; int min_dist = INF; void dfs1(int x, int step,int dist) {//最短期同样，取最短路径
    road1[step] = x; if (x == s) {//到达起点
        if (min_dist > dist) { min_dist = dist; r1.clear(); for (int i = step; i >= 0; i--)r1.push_back(road1[i]); } return; } for (int i = 0; i < prev_t[x].size(); i++) { int from = prev_t[x][i]; dfs1(from, step + 1,dist+Dist[from][x]); } } int main() { while (scanf("%d%d",&n,&m)!=EOF) { for (int i = 0; i < m;i++) { int from, to, one, L, T; scanf("%d%d%d%d%d", &from, &to, &one, &L, &T); G[from].push_back(edge(to, T, L)); Dist[from][to] = L; if (!one) { G[to].push_back(edge(from, T, L)); Dist[to][from] = L; } } scanf("%d%d",&s,&t); dijkstra1(s); dijkstra2(s); dfs1(t, 0, 0); dfs2(t, 0); if (r1 == r2) { printf("Time = %d; Distance = %d:",d_t[t],d_l[t]); for (int i = 0; i < r1.size(); i++) printf(" %d%s",r1[i],i+1==r1.size()? "\n" : " =>"); } else { printf("Time = %d:",d_t[t]); for (int i = 0; i < r1.size();i++) { printf(" %d%s", r1[i], i + 1 == r1.size() ? "\n" : " =>"); } printf("Distance = %d:", d_l[t]); for (int i = 0; i < r2.size();i++) { printf(" %d%s", r2[i], i + 1 == r2.size() ? "\n" : " =>"); } } } return 0; }

pat 团体天梯赛 L3-007. 天梯地图

L3-007. 天梯地图