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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#163540 | #7105. Pixel Art | oscaryang | AC ✓ | 246ms | 37000kb | C++14 | 2.5kb | 2023-09-04 10:28:36 | 2023-09-04 10:28:37 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define db double
#define mkp make_pair
#define pii pair<int,int>
#define vc vector
#define pb emplace_back
#define mem(a) memset(a,0,sizeof(a))
//#define int long long
const int N = 5e5+5, P = 1e9+7;
const int inf = 0x3f3f3f3f;
//const ll inf = 0x3f3f3f3f3f3f3f3f;
mt19937 gen(time(0));
//in&out
inline int read() {
int x = 0, w = 0; char ch = getchar(); while(!isdigit(ch)) w |= (ch=='-'), ch = getchar();
while(isdigit(ch)) x = (x*10)+(ch^48), ch=getchar(); return w?-x:x;
}
inline void write(int x) { if(x<0) putchar('-'), x = -x; if(x>9) write(x/10); putchar(x%10+'0'); }
inline void writec(int x) { write(x); putchar(32); }
inline void writee(int x) { write(x); putchar(10); }
//calc
inline void inc(int &x,int y) { x += y-P; x += (x>>31)&P; }
inline void dec(int &x,int y) { x -= y; x += (x>>31)&P; }
inline int pls(int x,int y) { x += y-P; x += (x>>31)&P; return x; }
inline void Max(int &x,int y) { if(x<y) x = y; }
inline void Min(int &x,int y) { if(x>y) x = y; }
inline int power(int a,int b) { int res = 1; for(;b;b >>= 1,a = 1ll*a*a%P) if(b&1) res = 1ll*res*a%P; return res; }
int n, m, q, tot;
ll sum, SUM;
struct node { int l, r, id; bool friend operator<(node A,node B) { return A.r<B.r; } };
vc<node> ad[N], dl[N];
set<node> S;
namespace dsu {
int fa[N];
inline int get(int x) { return x==fa[x]?x:fa[x] = get(fa[x]); }
inline void merge(int x,int y) { x = get(x); y = get(y); if(x!=y) fa[x] = y, --tot; }
}
using namespace dsu;
inline void solve() {
n = read(); m = read(); q = read();
S.clear(); tot = sum = 0; SUM = 0;
for(int i=1;i<=q;i++) fa[i] = i;
for(int i=1;i<=n;i++) ad[i].clear(), dl[i].clear();
for(int i=1;i<=q;i++) {
int r1 = read(), c1 = read(), r2 = read(), c2 = read();
ad[r1].pb(node{c1,c2,i}), dl[r2+1].pb(node{c1,c2,i});
}
S.insert(node{-1,-1,0}); S.insert(node{m+2,m+2,0});
for(int i=1;i<=n;i++) {
for(auto u:ad[i]) {
auto [l,r,id] = u; ++tot;
for(auto it = S.lower_bound(node{0,l,0});it!=S.end() && (*it).l<=r;++it)
merge(id,(*it).id);
}
for(auto u:dl[i]) S.erase(u), SUM -= u.r-u.l+1;
for(auto u:ad[i]) S.insert(u), SUM += u.r-u.l+1;
for(auto u:ad[i]) {
auto it = S.lower_bound(u); --it;
if((*it).r==u.l-1) merge((*it).id,u.id);
++it; ++it;
if((*it).l==u.r+1) merge((*it).id,u.id);
}
sum += SUM;
printf("%lld %d\n",sum,tot);
}
}
signed main() {
//freopen("a.in","r",stdin);
int t = read(); while(t--) solve();
return 0;
}
这程序好像有点Bug,我给组数据试试?
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 28636kb
input:
3 2 5 2 1 1 1 2 2 3 2 5 2 5 2 1 1 1 3 2 3 2 5 3 3 3 1 1 1 2 3 1 3 2 1 3 2 3
output:
2 1 5 2 3 1 6 1 3 1 4 1 6 2
result:
ok 7 lines
Test #2:
score: 0
Accepted
time: 246ms
memory: 37000kb
input:
2130 2 5 2 1 1 1 2 2 3 2 5 2 5 2 1 1 1 3 2 3 2 5 3 3 3 1 1 1 2 3 1 3 2 1 3 2 3 3 100 51 1 2 2 2 1 4 2 4 1 6 2 6 1 8 2 8 1 10 2 10 1 12 2 12 1 14 2 14 1 16 2 16 1 18 2 18 1 20 2 20 1 22 2 22 1 24 2 24 1 26 2 26 1 28 2 28 1 30 2 30 1 32 2 32 1 34 2 34 1 36 2 36 1 38 2 38 1 40 2 40 1 42 2 42 1 44 2 44 ...
output:
2 1 5 2 3 1 6 1 3 1 4 1 6 2 50 50 100 50 200 1 50 50 150 1 200 1 2 1 4 1 6 1 8 1 10 1 12 1 14 1 16 1 18 1 20 1 22 1 24 1 26 1 28 1 30 1 32 1 34 1 36 1 38 1 40 1 42 1 44 1 46 1 48 1 50 1 52 1 54 1 56 1 58 1 60 1 62 1 64 1 66 1 68 1 70 1 72 1 74 1 76 1 78 1 80 1 82 1 84 1 86 1 88 1 90 1 92 1 94 1 96 1...
result:
ok 355756 lines
Extra Test:
score: 0
Extra Test Passed