7-4 天梯地图 (30 分)
7-4 天梯地图 (30 分)
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
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <map>
#include <set>
#include <vector>
#include <cstring>
#include <queue>
#include <cmath>
#define FRER() freopen("in.txt","r",stdin);
#define FREW() freopen("out.txt","w",stdout);
#define mem(a,b) memset(a,b,sizeof(a));
#define go int T;scanf("%d",&T);for(int cas=1;cas<=T;cas++)
#define mod 1000000007
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
const int maxn = 1000000 + 7 , inf = 0x3f3f3f3f;
int nEdge,n,m,k,st,ed,sumT,sumW,resW,resT;
int totT,totW;
int head[maxn],nxt[maxn],to[maxn],t[maxn],w[maxn];
bool vis[maxn];
int d[maxn];
vector<int>prv1[maxn];
vector<int>prv2[maxn];
vector<int>path1;
vector<int>path2;
vector<int>_path;
void addEdge(int u,int v,int time,int weight){
to[nEdge] = v;
nxt[nEdge] = head[u];
t[nEdge] = time;
w[nEdge] = weight;
head[u] = nEdge++;
}
int gt[507][507];
int gw[507][507];
void dij(int c[],vector<int>prv[]){
for(int i=0;i<=n;i++) d[i] = inf;
d[st] = 0;
memset(vis, 0, sizeof(vis));
priority_queue<pii,vector<pii>,greater<pii> >q;
q.push(pii(d[st],st));
while(!q.empty()){
pii p = q.top();q.pop();
int u = p.second;
if(vis[u]) continue;
vis[u] = 1;
for(int e = head[u];~e;e=nxt[e]){
int v = to[e];
int cost = c[e];
if(d[v]>d[u]+cost){
d[v] = d[u]+cost;
prv[v].clear();
prv[v].push_back(u);
q.push(pii(d[v],v));
}
else if(d[v]==d[u]+cost){
prv[v].push_back(u);
}
}
}
}
void dfs1(int u){
if(u==st){
if(resW>sumW){
resW = sumW;
path1 = _path;
}
return;
}
for(int i=0;i<prv1[u].size();i++){
int v = prv1[u][i];
_path.push_back(v);
sumW+=gw[u][v];
dfs1(v);
sumW-=gw[u][v];
_path.erase(_path.end()-1);
}
}
void dfs2(int u){
if(u==st){
if(resT>sumT){
resT = sumT;
path2 = _path;
}
return;
}
for(int i=0;i<prv2[u].size();i++){
int v = prv2[u][i];
_path.push_back(v);
sumT++;
dfs2(v);
sumT--;
_path.erase(_path.end()-1);
}
}
int main(){
ios::sync_with_stdio(false);
cin>>n>>m;
nEdge = sumT = sumW = 0;
resT = resW = inf;
for(int i=0;i<=n;i++) head[i] = -1;
for(int i=1;i<=m;i++){
int u,v,tag,time,weight;
cin>>u>>v>>tag>>weight>>time;
if(!tag){
addEdge(u, v, time, weight);
addEdge(v, u, time, weight);
}else addEdge(u, v, time, weight);
gt[u][v] = gt[v][u] = time;
gw[u][v] = gw[v][u] = weight;
}
cin>>st>>ed;
dij(t,prv1);
totT = d[ed];
dij(w,prv2);
totW = d[ed];
_path.clear();
_path.push_back(ed);
dfs1(ed);
_path.clear();
_path.push_back(ed);
dfs2(ed);
if(path1==path2){
printf("Time = %d; Distance = %d: %d",totT,totW,st);
for(int i = (int)path1.size()-2;i>=0;i--)
printf(" => %d",path1[i]);
printf("\n");
}else{
printf("Time = %d: %d",totT,st);
for(int i = (int)path1.size()-2;i>=0;i--)
printf(" => %d",path1[i]);
printf("\n");
printf("Distance = %d: %d",totW,st);
for(int i = (int)path2.size()-2;i>=0;i--)
printf(" => %d",path2[i]);
printf("\n");
}
}