思路:一道模拟题,参考了大佬们的代码,目前尝试了bfs直接暴力,(把当前要扩展的链放到搜到的点上),80分,会超内存。
可以改成在询问时处理所有请求,并生成结果队列,能用引用的地方尽量用引用(不然很容易超时)。
还有对于题意,我刚开始有个误区,以为规则bi <= bi+1只有输入是a,b,c才有效,但是仔细读题,查询操作也是遵循这个规则的。
代码:bfs(之前似乎贴错代码啦。。重贴了一下。。)
80分:
#include <bits/stdc++.h>
#include <sstream>
using namespace std;
const int N = 505;
const int maxn = 4e5;
struct node {
int time;
vector <int> blk;
node(int a , vector <int> b) : time(a) , blk(b){
}
node() {
}
};
struct node1 {
int val , cg;
node1(int a , int b):val(a) , cg(b){}
};
vector <int> ans[N] , g[N];
vector <node> rq[N];
bool vis[N];
int t;
bool check(vector <int> &a , vector <int> &b) {
if(a.size() > b.size() || (a.size() == b.size() && a.back() < b.back())){
return true;
}
return false;
}
void solve(int a , int b) {
for(auto it = rq[a].begin() ; it != rq[a].end() ; ) {
int x = (*it).time;
if(x <= b) {
if(check((*it).blk , ans[a])) {
ans[a] = (*it).blk;
rq[a].erase(it);
continue;
}
}
it++;
}
}
void bfs(int a , int b , vector <int> c) {
memset(vis , 0 , sizeof(vis));
queue <node1> q;
q.push(node1(a , 0));
vis[a] = 1;
while(!q.empty()) {
node1 u = q.front();q.pop();
for(int i = 0 ; i < g[u.val].size() ; i++) {
int v = g[u.val][i];
if(!vis[v]) {
//cout << "size = " << c.size() << endl;
rq[v].push_back(node(b + (u.cg + 1) * t , c)); //传递
q.push(node1(v , u.cg + 1));
vis[v] = 1;
}
}
}
}
int main() {
int n , m;
ios::sync_with_stdio(0);
cin.tie(0);cout.tie(0);
cin >> n >> m;
for(int i = 0 ; i < m ; i++) {
int u , v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
int k;
cin >> t >> k;
for(int i = 1 ; i <= n ; i++) {
ans[i].push_back(0);
}
while(k--) {
int a , b , c;
cin >> a >> b;
if(cin.get() == '\n' || cin.eof()) {
solve(a , b);
cout << ans[a].size();
for(int i = 0 ; i < ans[a].size() ; i++) {
cout << " ";
cout << ans[a][i];
}
cout << "\n";
}
else {
cin >> c;
solve(a , b);
ans[a].push_back(c);
bfs(a , b , ans[a]);
}
}
return 0;
}
模拟法(100分):
#include <bits/stdc++.h>
using namespace std;
const int N = 505;
vector<vector<int>> g(505); //存储节点的边关系
vector<vector<int>> ans(505, {0}); //存储每个节点的主链
map <int , unordered_map<int , array < vector<int> , 2> > >ac;
int t , n , m , k;
bool check(vector <int> &a ,vector <int> &b) {
if(a.size() < b.size() || (a.size() == b.size() && a.back() > b.back())) {
return true;
}
return false;
}
void update(int time , int u) {
auto &cur = ans[u];
for(auto it : g[u]){
auto &tmp = ac[time][(it)][0];
if( ( check(tmp , cur) && !tmp.empty() ) || ( check(ans[it] , cur ) && tmp.empty() ) ){
tmp = cur;
}
}
}
void query(int time) {
for(auto& ac_d : ac) {
int ctime = ac_d.first;
if(ctime > time)break;
for(auto& ele : ac_d.second) {
int jd = ele.first;
auto & in = ele.second[0] , &out = ele.second[1];
bool flag = 0;
if(check(ans[jd] , in)) {
ans[jd] = in;
flag = 1;
}
for(int b : out) {
ans[jd].push_back(b);
}
if(flag || !out.empty()) {
update(ctime + t , jd);
}
}
}
ac.erase(ac.begin() , ac.upper_bound(time));
}
int main() {
ios::sync_with_stdio(false);
int n , m;
cin >> n >> m;
for(int i = 0 ; i < m ; i++) {
int u , v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
int k;
cin >> t >> k;
while(k--) {
int a , b , c;
cin >> a >> b;
if(cin.get() == '\n' || cin.eof()) {
query(b);
cout << ans[a].size();
for(int x : ans[a]) {
cout << " " << x;
}
cout << "\n";
}
else {
cin >> c;
ac[b][a][1].push_back(c);
}
}
return 0;
}