训练补题-牛客训练赛7
# 牛客训练赛7补题记录
# J: Pointer Analysis
题解的做法让我想起了spfa
但是似乎不是很正确？
是否有这样一种情况：
- 存在命令的某一排列，使得，但的任意排列都无法使
题解也没有给出证明。
AC代码
#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define Android ios::sync_with_stdio(false), cin.tie(NULL)
using namespace std;
const int maxn = 200 + 10;
struct statement{
int type;
int left, left_ptr, right, right_ptr;
bool operator <(const statement & s)const{
return type < s.type;
}
}st[maxn];
statement analysis(const string & left, const string & right){
statement ret = {0, 0, 0, 0, 0};
if(left.size() == 1 && right.size() == 1){
ret.left = left[0] - 'A';
if(islower(right[0])){ // A = x
ret.type = 1;
ret.right = right[0] - 'a';
}else{ // A = B
ret.type = 2;
ret.right = right[0] - 'A';
}
}else{
ret.left = left[0] - 'A';
ret.right = right[0] - 'A';
if(left.size() == 1){ // A = B.f
ret.type = 3;
ret.right_ptr = right[2] - 'a';
}else{ // A.f = B
ret.type = 4;
ret.left_ptr = left[2] - 'a';
}
}
return ret;
}
int start[5], finish[5];
struct pointer{
bool points_to[28];
}ptr[27];
struct instance{
pointer ptr[28];
}ins[28];
void solve(){
memset(start, 0x3f, sizeof start);
int n; cin >> n;
string _buffer1, _buffer2;
for(int i=0; i<n; i++){
cin >> _buffer1 >> _buffer2 >> _buffer2;
st[i] = analysis(_buffer1, _buffer2);
}
sort(st, st + n);
for(int i=0; i<n; i++){
finish[st[i].type] = i;
if(start[st[i].type] == 0x3f3f3f3f) start[st[i].type] = i;
}
vector<bool> inqueue(28, false);
queue<int> work;
for(int i=start[1]; i<=finish[1]; i++){
int a = st[i].left, o = st[i].right;
if(!ptr[a].points_to[o]){
ptr[a].points_to[o] = true;
work.push(a);
}
}
while(!work.empty()){
while(!work.empty()){
int u = work.front(); work.pop();
inqueue[u] = false;
for(int i=start[2]; i<=finish[2]; i++){
if(st[i].right != u) continue;
int l = st[i].left;
bool updated = false;
for(int k=0; k<26; k++){
if(!ptr[l].points_to[k] && ptr[u].points_to[k]){
updated = true;
ptr[l].points_to[k] = true;
}
}
if(updated && !inqueue[l]) {
work.push(l);
inqueue[l] = true;
}
}
}
for(int i=start[4]; i<=finish[4]; i++){
int l = st[i].left, lptr = st[i].left_ptr, r = st[i].right;
for(int j=0; j<26; j++){
if(!ptr[l].points_to[j]) continue;
for(int k=0; k<26; k++){
ins[j].ptr[lptr].points_to[k] |= ptr[r].points_to[k];
}
}
}
for(int i=start[3]; i<=finish[3]; i++){
int l = st[i].left, r = st[i].right, rptr = st[i].right_ptr;
bool updated = false;
for(int j=0; j<26; j++){
if(!ptr[r].points_to[j]) continue;
for(int k=0; k<26; k++){
if(!ptr[l].points_to[k] && ins[j].ptr[rptr].points_to[k]){
updated = true;
ptr[l].points_to[k] = true;
}
}
}
if(updated && !inqueue[l]){
work.push(l);
inqueue[l] = true;
}
}
}
for(int i=0; i<26; i++){
cout << (char)('A' + i) << ": ";
for(int j=0; j<26; j++){
if(ptr[i].points_to[j]) cout << (char)('a' + j);
}
cout << '\n';
}
}
signed main(){
Android;
solve();
}
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# A: Social Distancing
# C: A National Pandemic
# I: Valuable Forests
上次更新: 2021/02/24, 03:37:30