L3-007 天梯地图 (30 分)

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本题要求你实现一个天梯赛专属在线地图，队员输入自己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。

输入格式：
输入在第一行给出两个正整数N（2 ≤ N ≤ 500）和M，分别为地图中所有标记地点的个数和连接地点的道路条数。随后M行，每行按如下格式给出一条道路的信息：

V1 V2 one-way length time
其中V1和V2是道路的两个端点的编号（从0到N-1）；如果该道路是从V1到V2的单行线，则one-way为1，否则为0；length是道路的长度；time是通过该路所需要的时间。最后给出一对起点和终点的编号。

输出格式：
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：

Time = T: 起点 => 节点1 => … => 终点
然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：

Distance = D: 起点 => 节点1 => … => 终点
如果最快到达路线不唯一，则输出几条最快路线中最短的那条，题目保证这条路线是唯一的。而如果最短距离的路线不唯一，则输出途径节点数最少的那条，题目保证这条路线是唯一的。

如果这两条路线是完全一样的，则按下列格式输出：

Time = T; Distance = D: 起点 => 节点1 => … => 终点
输入样例1：
10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
5 4 0 2 3
5 9 1 1 4
0 6 0 1 1
7 3 1 1 2
8 3 1 1 2
2 5 0 2 2
2 1 1 1 1
1 5 0 1 3
1 4 0 1 1
9 7 1 1 3
3 1 0 2 5
6 3 1 2 1
5 3
输出样例1：
Time = 6: 5 => 4 => 8 => 3
Distance = 3: 5 => 1 => 3
输入样例2：
7 9
0 4 1 1 1
1 6 1 3 1
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 3 1
3 2 1 2 1
4 5 0 2 2
6 5 1 2 1
3 5
输出样例2：
Time = 3; Distance = 4: 3 => 2 => 5

两次Dijkstra即可，记录路径用数组模拟记录路径上每个点的父节点即可，用pre[v] = u来表示。判断节点数用cnt来记录每个点的节点数量即可。判断长度用dis数组记录长度即可。

复制代码

#include <bits/stdc++.h>
using namespace std;
int n , m , bg , ed , anstime , ansdist;
vector <int> ans1 , ans2;
struct node{
int u , w;
bool operator <(const node &a) const{
return a.w < w;
}
};
struct node2{
int u , w , time;
bool operator <(const node2 &a) const{
return a.time < time;
}
};
vector <node> mp1[500];
vector <node2> mp2[500];
void dijkstra1(){
int dis[500] , cnt[500] = {} , pre[500] = {};
priority_queue<node> q;
q.push({bg , 0});
for(int i = 0; i < 500; ++i){
dis[i] = 0x3f3f3f3f;
}
dis[bg] = 0;
while(!q.empty()){
node cur = q.top();
q.pop();
int dist = cur.w , u = cur.u;
if(dist != dis[u]){
continue;
}
for(int i = 0; i < mp1[u].size(); ++i){
int v = mp1[u][i].u , w = mp1[u][i].w;
if(dis[v] > dis[u] + w){
dis[v] = dis[u] + w;
q.push({v , dis[v]});
cnt[v] = cnt[u] + 1;
pre[v] = u;
}
else if(dis[v] == dis[u] + w && cnt[u]+1 < cnt[v]){
cnt[v] = cnt[u] + 1;
pre[v] = u;
}
}
}
int bound = ed;
while(bound != bg){
ans1.push_back(bound);
bound = pre[bound];
}
ans1.push_back(bound);
ansdist = dis[ed];
}
void dijkstra2(){
int tim[500] , pre[500] = {} , dis[500] = {};
priority_queue<node2> q;
q.push({bg , 0 , 0});
for(int i = 0; i < 500; ++i){
tim[i] = 0x3f3f3f3f;
}
tim[bg] = 0;
while(!q.empty()){
node2 cur = q.top();
q.pop();
int timt = cur.time , u = cur.u , dist = cur.w;
if(timt != tim[u]){
continue;
}
for(int i = 0; i < mp2[u].size(); ++i){
int v = mp2[u][i].u , time = mp2[u][i].time , w = mp2[u][i].w;
if(tim[v] > tim[u] + time){
tim[v] = tim[u] + time;
dis[v] = dist+w;
q.push({v , dist+w , tim[v]});
pre[v] = u;
}
else if(tim[v] == tim[u] + time && dis[u]+w < dis[v]){
dis[v] = dis[u] + w;
pre[v] = u;
}
}
}
int bound = ed;
while(bound != bg){
ans2.push_back(bound);
bound = pre[bound];
}
ans2.push_back(bound);
anstime = tim[ed];
}
int main(){
cin >> n >> m;
for(int i = 0; i < m; ++i){
int v1 , v2 , one_way , len , time;
cin >> v1 >> v2 >> one_way >> len >> time;
mp1[v1].push_back({v2 , len});
mp2[v1].push_back({v2 , len , time});
if(one_way == 0){
mp1[v2].push_back({v1 , len});
mp2[v2].push_back({v1 , len , time});
}
}
cin >> bg >> ed;
dijkstra1();
dijkstra2();
if(ans1 == ans2){
cout << "Time = " << anstime << "; Distance = " << ansdist << ": ";
for(int i = ans2.size()-1; i >= 0; --i){
cout << ans2[i] << (i == 0 ? "" : " => ");
}
}
else{
cout << "Time = " << anstime <<": ";
for(int i = ans2.size()-1; i >= 0; --i){
cout << ans2[i] << (i == 0 ? "" : " => ");
}
cout << endl;
cout << "Distance = " << ansdist <<": ";
for(int i = ans1.size()-1; i >= 0; --i){
cout << ans1[i] << (i == 0 ? "" : " => ");
}
}
return 0;
}

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#include <bits/stdc++.h>
using namespace std;
int n , m , bg , ed , anstime , ansdist;
vector <int> ans1 , ans2;
struct node{
int u , w;
bool operator <(const node &a) const{
return a.w < w;
}
};
struct node2{
int u , w , time;
bool operator <(const node2 &a) const{
return a.time < time;
}
};
vector <node> mp1[500];
vector <node2> mp2[500];
void dijkstra1(){
int dis[500] , cnt[500] = {} , pre[500] = {};
priority_queue<node> q;
q.push({bg , 0});
for(int i = 0; i < 500; ++i){
dis[i] = 0x3f3f3f3f;
}
dis[bg] = 0;
while(!q.empty()){
node cur = q.top();
q.pop();
int dist = cur.w , u = cur.u;
if(dist != dis[u]){
continue;
}
for(int i = 0; i < mp1[u].size(); ++i){
int v = mp1[u][i].u , w = mp1[u][i].w;
if(dis[v] > dis[u] + w){
dis[v] = dis[u] + w;
q.push({v , dis[v]});
cnt[v] = cnt[u] + 1;
pre[v] = u;
}
else if(dis[v] == dis[u] + w && cnt[u]+1 < cnt[v]){
cnt[v] = cnt[u] + 1;
pre[v] = u;
}
}
}
int bound = ed;
while(bound != bg){
ans1.push_back(bound);
bound = pre[bound];
}
ans1.push_back(bound);
ansdist = dis[ed];
}
void dijkstra2(){
int tim[500] , pre[500] = {} , dis[500] = {};
priority_queue<node2> q;
q.push({bg , 0 , 0});
for(int i = 0; i < 500; ++i){
tim[i] = 0x3f3f3f3f;
}
tim[bg] = 0;
while(!q.empty()){
node2 cur = q.top();
q.pop();
int timt = cur.time , u = cur.u , dist = cur.w;
if(timt != tim[u]){
continue;
}
for(int i = 0; i < mp2[u].size(); ++i){
int v = mp2[u][i].u , time = mp2[u][i].time , w = mp2[u][i].w;
if(tim[v] > tim[u] + time){
tim[v] = tim[u] + time;
dis[v] = dist+w;
q.push({v , dist+w , tim[v]});
pre[v] = u;
}
else if(tim[v] == tim[u] + time && dis[u]+w < dis[v]){
dis[v] = dis[u] + w;
pre[v] = u;
}
}
}
int bound = ed;
while(bound != bg){
ans2.push_back(bound);
bound = pre[bound];
}
ans2.push_back(bound);
anstime = tim[ed];
}
int main(){
cin >> n >> m;
for(int i = 0; i < m; ++i){
int v1 , v2 , one_way , len , time;
cin >> v1 >> v2 >> one_way >> len >> time;
mp1[v1].push_back({v2 , len});
mp2[v1].push_back({v2 , len , time});
if(one_way == 0){
mp1[v2].push_back({v1 , len});
mp2[v2].push_back({v1 , len , time});
}
}
cin >> bg >> ed;
dijkstra1();
dijkstra2();
if(ans1 == ans2){
cout << "Time = " << anstime << "; Distance = " << ansdist << ": ";
for(int i = ans2.size()-1; i >= 0; --i){
cout << ans2[i] << (i == 0 ? "" : " => ");
}
}
else{
cout << "Time = " << anstime <<": ";
for(int i = ans2.size()-1; i >= 0; --i){
cout << ans2[i] << (i == 0 ? "" : " => ");
}
cout << endl;
cout << "Distance = " << ansdist <<": ";
for(int i = ans1.size()-1; i >= 0; --i){
cout << ans1[i] << (i == 0 ? "" : " => ");
}
}
return 0;
}