2022NOIPA层联测34
A. bs 串
只知道去找环然后挨个判断……正解是把不同色的边连上，枚举哪两个同色的边两端已经联通。二分+并查集。
code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 3;
const int M = 4e5 + 3;
int n, m, q, f[maxn], fa[maxn], tot;
char s[maxn];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct node
{
int u, v;
}e[maxn], a[maxn], b[maxn];
int find(int x) {return f[x] == x ? x : f[x] = find(f[x]);}
int find2(int x) {return fa[x] == x ? x : fa[x] = find2(fa[x]);}
bool check()
{
for(int i=1; i<=tot; i++)
{
if(find(e[i].u) == find(e[i].v)) return 1;
}
return 0;
}
bool check2(int mid)
{
memcpy(fa, f, sizeof(f));
for(int i=1; i<=mid; i++)
{
if(!b[i].u) continue;
fa[find2(b[i].u)] = find2(b[i].v);
}
for(int i=1; i<=tot; i++)
{
if(find2(e[i].u) == find2(e[i].v)) return 1;
}
for(int i=1; i<=mid; i++)
{
if(!a[i].u) continue;
if(find2(a[i].u) == find2(a[i].v)) return 1;
}
return 0;
}
int main()
{
freopen("bssb.in", "r", stdin);
freopen("bssb.out", "w", stdout);
n = read(), m = read(), q = read();
scanf("%s", s+1);
for(int i=1; i<=n; i++) f[i] = i;
for(int i=1; i<=m; i++)
{
int u = read(), v = read();
if(s[u] == s[v]) e[++tot] = (node){u, v};
else f[find(u)] = find(v);
}
if(check())
{
for(int i=1; i<=q; i++)
{
printf("Yes\n");
}
exit(0);
}
for(int i=1; i<=q; i++)
{
int u = read(), v = read();
if(s[u] == s[v]) a[i] = (node){u, v};
else b[i] = (node){u, v};
}
int l = 1, r = q + 1;
while(l < r)
{
int mid = (l + r) >> 1;
if(check2(mid)) r = mid;
else l = mid + 1;
}
for(int i=1; i<l; i++) printf("No\n");
for(int i=l; i<=q; i++) printf("Yes\n");
return 0;
}
B. 英语作文
题目告诉我顺序排列，我想排列当然是有顺序的这不是废话吗……然而顺序排列的意思是按单词给出的顺序排列……连样例都没过就乱写……
然而由于a是起点以a开头随时都可以，第一个式子应该写成：
f[i][0] = (f[i-1][0]+1ll*b[i][0]*(f[i-1][1]+f[i-1][2]+1)%mod)%mod;至于加了个f[i-1][a/b/c]不是代表可以连续的拼同一种词性，它代表直接继承，i 可以不选嘛。
code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e5 + 3;
const int mod = 998244353;
int f[maxn][4], n;
bool b[maxn][4];
char s[4];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
int main()
{
freopen("paper.in", "r", stdin);
freopen("paper.out", "w", stdout);
n = read(); int fl = 0;
for(int i=1; i<=n; i++)
{
scanf("%s", s);
if(s[0] == 'T')
{
b[i][0] = 1;
if(!fl) fl = i;
}
if(s[1] == 'T') b[i][1] = 1;
if(s[2] == 'T') b[i][2] = 1;
}
for(int i=fl; i<=n; i++)
{
f[i][0] = (f[i-1][0]+1ll*b[i][0]*(f[i-1][1]+f[i-1][2]+1)%mod)%mod;
f[i][1] = (f[i-1][1]+1ll*b[i][1]*(f[i-1][0]+f[i-1][2])%mod)%mod;
f[i][2] = (f[i-1][2]+1ll*b[i][2]*f[i-1][1]%mod)%mod;
}
printf("%d\n", f[n][2]);
return 0;
}
D. 愤怒的小鸟
一看就是反悔贪心然而不知道反悔贪心怎么用……于是%%%Chen_jr
具体原理还不是很理解感觉应该去研究一下费用流……
合法的反悔方案似乎很有规律，顺序毫无关系因为所有的方案都可以被反悔：
- 3 2 1
3：直接拿3
2：拿2 、3换2
1：拿1、拿2+2换1、拿3+3换1、拿2+2换3+3换1、拿3+3换2+2换1
code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 3;
int n, p, q, r, a[maxn], b[maxn], c[maxn], vis[maxn];
typedef pair<ll, int> pli;
priority_queue<pli> qcb, qca, qbc, qba;
priority_queue<pli, vector<pli>, greater<pli> > qa, qb, qc;
ll ans;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void make_3star(int x)
{
vis[x] = 3;
qcb.push(pli(c[x]-b[x], x));
qca.push(pli(c[x]-a[x], x));
}
void make_2star(int x)
{
vis[x] = 2;
qbc.push(pli(b[x]-c[x], x));
qba.push(pli(b[x]-a[x], x));
}
void make_1star(int x)
{
vis[x] = 1;
}
ll get_1star()
{
while(!qa.empty() && vis[qa.top().second]) qa.pop();
return qa.empty() ? 1e18 : qa.top().first;
}
ll get_2star()
{
while(!qb.empty() && vis[qb.top().second]) qb.pop();
return qb.empty() ? 1e18 : qb.top().first;
}
ll get_3star()
{
while(!qc.empty() && vis[qc.top().second]) qc.pop();
return qc.empty() ? 1e18 : qc.top().first;
}
ll make3to2()
{
while(!qcb.empty() && vis[qcb.top().second] != 3) qcb.pop();
return qcb.empty() ? -1e18 : qcb.top().first;
}
ll make3to1()
{
while(!qca.empty() && vis[qca.top().second] != 3) qca.pop();
return qca.empty() ? -1e18 : qca.top().first;
}
ll make2to3()
{
while(!qbc.empty() && vis[qbc.top().second] != 2) qbc.pop();
return qbc.empty() ? -1e18 : qbc.top().first;
}
ll make2to1()
{
while(!qba.empty() && vis[qba.top().second] != 2) qba.pop();
return qba.empty() ? -1e18 : qba.top().first;
}
int main()
{
freopen("angrybird.in", "r", stdin);
freopen("angrybird.out", "w", stdout);
n = read(), p = read(), q = read(), r = read();
for(int i=1; i<=n; i++) a[i] = read();
for(int i=1; i<=n; i++) b[i] = read();
for(int i=1; i<=n; i++) c[i] = read();
for(int i=1; i<=n; i++) b[i] = min(b[i], c[i]);
for(int i=1; i<=n; i++) a[i] = min(a[i], b[i]);
for(int i=1; i<=n; i++)
{
qa.push(pli(a[i], i)); qb.push(pli(b[i], i)); qc.push(pli(c[i], i));
}
p = p - max(q, r); q = q - r;
for(int i=1; i<=r; i++)
{
make_3star(qc.top().second); ans += qc.top().first; qc.pop();
}
for(int i=1; i<=q; i++)
{
ll v1 = get_2star(), v2 = get_3star() - make3to2();
if(v1 < v2)
{
ans += v1;
int x = qb.top().second; make_2star(x); qb.pop();
}
else
{
ans += v2;
int x = qc.top().second, y = qcb.top().second;
make_3star(x); make_2star(y);
}
}
for(int i=1; i<=p; i++)
{
ll v1 = get_1star(), v2 = get_2star()-make2to1(), v3 = get_3star()-make3to1();
ll v4 = get_2star()-make2to3()-make3to1(), v5 = get_3star()-make3to2()-make2to1();
if(v1 < v2 && v1 < v3 && v1 < v4 && v1 < v5)
{
ans += v1;
int x = qa.top().second; qa.pop();
make_1star(x);
}
else if(v2 < v3 && v2 < v4 && v2 < v5)
{
ans += v2;
int x = qb.top().second, y = qba.top().second;
qb.pop(), qba.pop();
make_2star(x), make_1star(y);
}
else if(v3 < v4 && v3 < v5)
{
ans += v3;
int x = qc.top().second, y = qca.top().second;
qc.pop(), qca.pop();
make_3star(x), make_1star(y);
}
else if(v4 < v5)
{
ans += v4;
int x = qb.top().second, y = qbc.top().second, z = qca.top().second;
qb.pop(); qbc.pop(); qca.pop();
make_2star(x), make_3star(y), make_1star(z);
}
else
{
ans += v5;
int x = qc.top().second, y = qcb.top().second, z = qba.top().second;
qc.pop(); qcb.pop(); qba.pop();
make_3star(x), make_2star(y), make_1star(z);
}
}
printf("%lld\n", ans);
return 0;
}- 1 2 3
1：直接拿1
2：拿2、1换2
3：拿3、拿2+2换3、拿1+1换3、拿2+2换1+1换3、拿1+1换2+2换3
code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 3;
int n, p, q, r, a[maxn], b[maxn], c[maxn], vis[maxn];
typedef pair<ll, int> pli;
priority_queue<pli> qab, qac, qba, qbc;
priority_queue<pli, vector<pli>, greater<pli> > qa, qb, qc;
ll ans;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void make_1star(int x)
{
vis[x] = 1;
qab.push(pli(a[x]-b[x], x));
qac.push(pli(a[x]-c[x], x));
}
void make_2star(int x)
{
vis[x] = 2;
qba.push(pli(b[x]-a[x], x));
qbc.push(pli(b[x]-c[x], x));
}
void make_3star(int x)
{
vis[x] = 3;
}
ll get_1star()
{
while(!qa.empty() && vis[qa.top().second]) qa.pop();
return qa.empty() ? 1e18 : qa.top().first;
}
ll get_2star()
{
while(!qb.empty() && vis[qb.top().second]) qb.pop();
return qb.empty() ? 1e18 : qb.top().first;
}
ll get_3star()
{
while(!qc.empty() && vis[qc.top().second]) qc.pop();
return qc.empty() ? 1e18 : qc.top().first;
}
ll make1to2()
{
while(!qab.empty() && vis[qab.top().second] != 1) qab.pop();
return qab.empty() ? -1e18 : qab.top().first;
}
ll make1to3()
{
while(!qac.empty() && vis[qac.top().second] != 1) qac.pop();
return qac.empty() ? -1e18 : qac.top().first;
}
ll make2to1()
{
while(!qba.empty() && vis[qba.top().second] != 2) qba.pop();
return qba.empty() ? -1e18 : qba.top().first;
}
ll make2to3()
{
while(!qbc.empty() && vis[qbc.top().second] != 2) qbc.pop();
return qbc.empty() ? -1e18 : qbc.top().first;
}
int main()
{
freopen("angrybird.in", "r", stdin);
freopen("angrybird.out", "w", stdout);
n = read(), p = read(), q = read(), r = read();
for(int i=1; i<=n; i++) a[i] = read();
for(int i=1; i<=n; i++) b[i] = read();
for(int i=1; i<=n; i++) c[i] = read();
for(int i=1; i<=n; i++) b[i] = min(b[i], c[i]);
for(int i=1; i<=n; i++) a[i] = min(a[i], b[i]);
p = p - max(q, r); q = q - r;
for(int i=1; i<=n; i++)
{
qa.push(pli(a[i], i)); qb.push(pli(b[i], i)); qc.push(pli(c[i], i));
}
for(int i=1; i<=p; i++)
{
make_1star(qa.top().second); ans += qa.top().first; qa.pop();
}
for(int i=1; i<=q; i++)
{
ll v1 = get_2star(), v2 = get_1star() - make1to2();
if(v1 < v2)
{
ans += v1;
int x = qb.top().second; make_2star(x); qb.pop();
}
else
{
ans += v2;
int x = qa.top().second, y = qab.top().second;
make_1star(x), make_2star(y);
}
}
for(int i=1; i<=r; i++)
{
ll v1 = get_3star(), v2 = get_2star()-make2to3(), v3 = get_1star()-make1to3();
ll v4 = get_2star()-make2to1()-make1to3(), v5 = get_1star()-make1to2()-make2to3();
if(v1 < v2 && v1 < v3 && v1 < v4 && v1 < v5)
{
ans += v1;
int x = qc.top().second; qc.pop();
make_3star(x);
}
else if(v2 < v3 && v2 < v4 && v2 < v5)
{
ans += v2;
int x = qb.top().second, y = qbc.top().second;
qb.pop(), qbc.pop();
make_2star(x), make_3star(y);
}
else if(v3 < v4 && v3 < v5)
{
ans += v3;
int x = qa.top().second, y = qac.top().second;
qa.pop(), qac.pop();
make_1star(x), make_3star(y);
}
else if(v4 < v5)
{
ans += v4;
int x = qb.top().second, y = qba.top().second, z = qac.top().second;
qb.pop(), qba.pop(), qac.pop();
make_2star(x), make_1star(y), make_3star(z);
}
else
{
ans += v5;
int x = qa.top().second, y = qab.top().second, z = qbc.top().second;
qa.pop(), qab.pop(), qbc.pop();
make_1star(x), make_2star(y), make_3star(z);
}
}
printf("%lld\n", ans);
return 0;
}包括好像比较显然，加号减号的方向和堆的种类还有感觉有一种更好理解的求pq的方法：q = q - r, p = p - q - r都是对的。
code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 3;
int n, p, q, r, a[maxn], b[maxn], c[maxn], vis[maxn];
typedef pair<ll, int> pli;
//priority_queue<pli> qab, qac, qba, qbc;
priority_queue<pli, vector<pli>, greater<pli> > qa, qb, qc, qab, qac, qba, qbc;
ll ans;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
void make_1star(int x)
{
vis[x] = 1;
qab.push(pli(b[x]-a[x], x));
qac.push(pli(c[x]-a[x], x));
}
void make_2star(int x)
{
vis[x] = 2;
qba.push(pli(a[x]-b[x], x));
qbc.push(pli(c[x]-b[x], x));
}
void make_3star(int x)
{
vis[x] = 3;
}
ll get_1star()
{
while(!qa.empty() && vis[qa.top().second]) qa.pop();
return qa.empty() ? 1e18 : qa.top().first;
}
ll get_2star()
{
while(!qb.empty() && vis[qb.top().second]) qb.pop();
return qb.empty() ? 1e18 : qb.top().first;
}
ll get_3star()
{
while(!qc.empty() && vis[qc.top().second]) qc.pop();
return qc.empty() ? 1e18 : qc.top().first;
}
ll make1to2()
{
while(!qab.empty() && vis[qab.top().second] != 1) qab.pop();
return qab.empty() ? 1e18 : qab.top().first;
}
ll make1to3()
{
while(!qac.empty() && vis[qac.top().second] != 1) qac.pop();
return qac.empty() ? 1e18 : qac.top().first;
}
ll make2to1()
{
while(!qba.empty() && vis[qba.top().second] != 2) qba.pop();
return qba.empty() ? 1e18 : qba.top().first;
}
ll make2to3()
{
while(!qbc.empty() && vis[qbc.top().second] != 2) qbc.pop();
return qbc.empty() ? 1e18 : qbc.top().first;
}
int main()
{
freopen("angrybird.in", "r", stdin);
freopen("angrybird.out", "w", stdout);
n = read(), p = read(), q = read(), r = read();
for(int i=1; i<=n; i++) a[i] = read();
for(int i=1; i<=n; i++) b[i] = read();
for(int i=1; i<=n; i++) c[i] = read();
for(int i=1; i<=n; i++) b[i] = min(b[i], c[i]);
for(int i=1; i<=n; i++) a[i] = min(a[i], b[i]);
q = q - r; p = p - q - r;
for(int i=1; i<=n; i++)
{
qa.push(pli(a[i], i)); qb.push(pli(b[i], i)); qc.push(pli(c[i], i));
}
for(int i=1; i<=p; i++)
{
make_1star(qa.top().second); ans += qa.top().first; qa.pop();
}
for(int i=1; i<=q; i++)
{
ll v1 = get_2star(), v2 = get_1star() + make1to2();
if(v1 < v2)
{
ans += v1;
int x = qb.top().second; make_2star(x); qb.pop();
}
else
{
ans += v2;
int x = qa.top().second, y = qab.top().second;
make_1star(x), make_2star(y);
}
}
for(int i=1; i<=r; i++)
{
ll v1 = get_3star(), v2 = get_2star()+make2to3(), v3 = get_1star()+make1to3();
ll v4 = get_2star()+make2to1()+make1to3(), v5 = get_1star()+make1to2()+make2to3();
if(v1 < v2 && v1 < v3 && v1 < v4 && v1 < v5)
{
ans += v1;
int x = qc.top().second; qc.pop();
make_3star(x);
}
else if(v2 < v3 && v2 < v4 && v2 < v5)
{
ans += v2;
int x = qb.top().second, y = qbc.top().second;
qb.pop(), qbc.pop();
make_2star(x), make_3star(y);
}
else if(v3 < v4 && v3 < v5)
{
ans += v3;
int x = qa.top().second, y = qac.top().second;
qa.pop(), qac.pop();
make_1star(x), make_3star(y);
}
else if(v4 < v5)
{
ans += v4;
int x = qb.top().second, y = qba.top().second, z = qac.top().second;
qb.pop(), qba.pop(), qac.pop();
make_2star(x), make_1star(y), make_3star(z);
}
else
{
ans += v5;
int x = qa.top().second, y = qab.top().second, z = qbc.top().second;
qa.pop(), qab.pop(), qbc.pop();
make_1star(x), make_2star(y), make_3star(z);
}
}
printf("%lld\n", ans);
return 0;
}
时光花火，水月星辰
