ARC065
C - 白昼夢 / Daydream
直接递推就好
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
string s;
int dp[MAXN];
void Solve() {
cin >> s;
dp[0] = 1;
for(int i = 1 ; i <= s.length() ; ++i) {
if(i >= 5) {
dp[i] |= (dp[i - 5] && s.substr(i - 5,5) == "dream");
dp[i] |= (dp[i - 5] && s.substr(i - 5,5) == "erase");
}
if(i >= 6) {
dp[i] |= (dp[i - 6] && s.substr(i - 6,6) == "eraser");
}
if(i >= 7) {
dp[i] |= (dp[i - 7] && s.substr(i - 7,7) == "dreamer");
}
}
if(dp[s.length()]) puts("YES");
else puts("NO");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
D - 連結 / Connectivity
把图两次染色，然后标号，会得到一个点对，这个点对相同的点才在两个图里都连通
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
map<pii,int> zz;
int N,K,L;
int fa[MAXN],col[MAXN][2],tot;
int getfa(int x) {
return fa[x] == x ? x : fa[x] = getfa(fa[x]);
}
void Solve() {
read(N);read(K);read(L);
for(int i = 1 ; i <= N ; ++i) fa[i] = i;
int a,b;
for(int i = 1 ; i <= K ; ++i) {
read(a);read(b);
fa[getfa(a)] = getfa(b);
}
tot = 0;
for(int i = 1 ; i <= N ; ++i) {
if(getfa(i) == i) col[i][0] = ++tot;
}
for(int i = 1 ; i <= N ; ++i) col[i][0] = col[getfa(i)][0];
for(int i = 1 ; i <= N ; ++i) fa[i] = i;
for(int i = 1 ; i <= L ; ++i) {
read(a);read(b);
fa[getfa(a)] = getfa(b);
}
tot = 0;
for(int i = 1 ; i <= N ; ++i) {
if(getfa(i) == i) col[i][1] = ++tot;
}
for(int i = 1 ; i <= N ; ++i) col[i][1] = col[getfa(i)][1];
for(int i = 1 ; i <= N ; ++i) {
zz[mp(col[i][0],col[i][1])]++;
}
for(int i = 1 ; i <= N ; ++i) {
out(zz[mp(col[i][0],col[i][1])]);space;
}
enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
E - へんなコンパス / Manhattan Compass
曼哈顿转切比雪夫
然后把点对排序，每个点用横坐标相连就是
(x_{a} - x_{b} = D)
并且$ -D<= y_{a} - y_{b} <= D$
这是一段区间，我们可以把区间里的点两两相连，再和a点相连，是等价的
然后记录每个点连出去的边有多少，和初始的两个点连通的所有点之间的边数加起来就是答案
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct node {
int to,next;
}E[MAXN * 10];
int head[MAXN],sumE;
int N,a,b;
int x[MAXN],y[MAXN],D,id[MAXN],cnt[MAXN];
pii p[MAXN],tmp[MAXN];
int st[MAXN],ed[MAXN];
int fa[MAXN];
int getfa(int x) {
return fa[x] == x ? x : fa[x] = getfa(fa[x]);
}
void Merge(int x,int y) {
fa[getfa(x)] = getfa(y);
}
void AddE(int on) {
for(int i = 1 ; i <= N ; ++i) {
id[i] = i;
}
sort(id + 1,id + N + 1,[](int a,int b){return p[a] < p[b];});
memset(st,0,sizeof(st));memset(ed,0,sizeof(ed));
for(int i = 1 ; i <= N ; ++i) {
tmp[i] = p[id[i]];
}
for(int i = 1 ; i <= N ; ++i) {
int l = lower_bound(tmp + 1,tmp + N + 1,mp(tmp[i].fi + D,tmp[i].se - D + on)) - tmp;
int r = upper_bound(tmp + 1,tmp + N + 1,mp(tmp[i].fi + D,tmp[i].se + D - on)) - tmp - 1;
if(l <= r) {
st[l]++;ed[r]++;
Merge(id[i],id[l]);
cnt[id[i]] += r - l + 1;
}
}
int pre = 0;
for(int i = 1 ; i < N ; ++i) {
pre += st[i];pre -= ed[i];
if(pre) Merge(id[i],id[i + 1]);
}
}
void Solve() {
read(N);read(a);read(b);
for(int i = 1 ; i <= N ; ++i) {
read(x[i]);read(y[i]);
fa[i] = i;
}
D = abs(x[a] - x[b]) + abs(y[a] - y[b]);
for(int i = 1 ; i <= N ; ++i) p[i] = mp(x[i] + y[i],x[i] - y[i]);
AddE(0);
for(int i = 1 ; i <= N ; ++i) swap(p[i].fi,p[i].se);
AddE(1);
int64 ans = 0;
for(int i = 1 ; i <= N ; ++i) {
if(getfa(i) == getfa(a)) ans += cnt[i];
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
F - シャッフル / Shuffling
就是对于一个区间，记录能用a个0和b个1
然后对于某个位置填0还是填1
由于能用的数是总和是确定的，我们只要记录当前有a个0没用就可以了
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 3005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007;
int ri[MAXN],f[2][MAXN];
int N,M;
char s[MAXN];
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void update(int &x,int y) {
x = inc(x,y);
}
void Solve() {
read(N);read(M);
scanf("%s",s + 1);
int l,r;
for(int i = 1 ; i <= M ; ++i) {
read(l);read(r);
ri[l] = max(ri[l],r);
}
int p = 0,all = 0;
int cur = 0;
f[cur][0] = 1;
for(int i = 1 ; i <= N ; ++i) {
if(p < i - 1) {p = i - 1;}
if(ri[i] > p) {
all += ri[i] - p;
int a = 0,b = 0;
for(int j = p + 1 ; j <= ri[i] ; ++j) {
if(s[j] == '0') ++a;
else ++b;
}
p = ri[i];
memset(f[cur ^ 1],0,sizeof(f[cur ^ 1]));
for(int j = a ; j <= N ; ++j) f[cur ^ 1][j] = f[cur][j - a];
cur ^= 1;
}
if(!all) continue;
memset(f[cur ^ 1],0,sizeof(f[cur ^ 1]));
for(int j = 0 ; j <= all ; ++j) {
if(j < all) update(f[cur ^ 1][j],f[cur][j]);
if(j >= 1) update(f[cur ^ 1][j - 1],f[cur][j]);
}
--all;
cur ^= 1;
}
out(f[cur][0]);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}