本来以为是1199rated的。。仔细一看发现是1999，所以就做了一下
这场涨分很轻松啊。。。为啥又没打
等pkuwc考完我一定打一场atcoder（咕咕咕，咕咕咕，咕咕咕咕咕咕咕~）
但是其实我思维速度上真的有点不行。。。
A - Bulletin Board
输出\((N - W + 1)(N - H + 1)\)
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 20000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,H,W;
void Solve() {
read(N);read(H);read(W);
out(1LL * (N - H + 1) * (N - W + 1));enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
} B - Contests
统计这三个区间的个数，输出最小值
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 20000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
int P[105],A,B;
int cnt[3];
void Solve() {
read(N);read(A);read(B);
for(int i = 1 ; i <= N ; ++i) {
read(P[i]);
if(P[i] <= A) ++cnt[0];
else if(P[i] <= B) ++cnt[1];
else cnt[2]++;
}
out(min(min(cnt[0],cnt[1]),cnt[2]));enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
} C - Alternating Path
相邻的黑白格子之间有边，答案是每个联通块里的黑白点个数乘积之和
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 20000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int H,W,a[405][405],cnt[2];
char s[405][405];
bool vis[405][405];
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
bool in_range(int x,int y) {
return x >= 1 && x <= H && y >= 1 && y <= W;
}
void dfs(int x,int y) {
cnt[a[x][y]]++;vis[x][y] = 1;
for(int k = 0 ; k < 4 ; ++k) {
int tx = x + dx[k],ty = y + dy[k];
if(in_range(tx,ty)) {
if(!vis[tx][ty] && a[tx][ty] != a[x][y]) {
dfs(tx,ty);
}
}
}
}
void Solve() {
read(H);read(W);
for(int i = 1 ; i <= H ; i++) {
scanf("%s",s[i] + 1);
for(int j = 1 ; j <= W ; ++j) {
if(s[i][j] == '#') a[i][j] = 1;
else a[i][j] = 0;
}
}
int64 ans = 0;
for(int i = 1 ; i <= H ; ++i) {
for(int j = 1 ; j <= W ; ++j) {
if(!vis[i][j]) {
cnt[1] = 0;cnt[0] = 0;
dfs(i,j);
ans += 1LL * cnt[1] * cnt[0];
}
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
} D - Nearest Card Game
先二分一个k，使得\([x - k,x + k]\)区间里的数的个数小于等于\((x + k,\infty)\)里的数，且\(k\)最大，先把\([x - k,x + k]\)这些区间里这么多数两个人分别取走
然后从这个状态暴力模拟（大于x的数不会超过两个了），使得所有的数都小于x，剩下的就是交替一个一个取了，可以用一个前缀和预处理出来
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,Q;
int64 A[MAXN],sum[MAXN],b[MAXN];
pii Calc(int64 l,int64 r) {
int a = lower_bound(A + 1,A + N + 1,l) - A;
int b = upper_bound(A + 1,A + N + 1,r) - A - 1;
return mp(a,b);
}
int64 Process(int64 x) {
if(x >= A[N]) return sum[N];
int64 L = 0,R = A[N];
int p = lower_bound(A + 1,A + N + 1,x) - A;
while(L < R) {
int64 mid = (L + R + 1) >> 1;
pii t = Calc(x - mid,x + mid);
int rem = N - p + 1 - (t.se - p + 1);
if(rem >= t.se - t.fi + 1) L = mid;
else R = mid - 1;
}
pii t = Calc(x - L,x + L);
int len = t.se - t.fi + 1;
int64 res = b[N] - b[N - len];
int q = N - len,h0 = t.fi - 1,h1 = t.se + 1;
while(q >= h1) {
res += A[q];--q;
if(h1 > q && h0 < 1) break;
int now;
if(h1 > q) now = h0;
else if(h0 < 1) now = h1;
else {
if(x - A[h0] <= A[h1] - x) now = h0;
else now = h1;
}
if(now == h1) ++h1;
if(now == h0) --h0;
}
res += sum[h0];
return res;
}
void Solve() {
read(N);read(Q);
for(int i = 1 ; i <= N ; ++i) read(A[i]);
for(int i = 1 ; i <= N ; ++i) {
sum[i] = A[i];
if(i >= 2) sum[i] += sum[i - 2];
b[i] = A[i] + b[i - 1];
}
int64 x;
for(int i = 1 ; i <= Q ; ++i) {
read(x);
out(Process(x));enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
} E - Attack to a Tree
这个直接dp就好了，\(dp[u][j][0/1]\)表示以\(u\)为根，砍断了\(j\)条边，0表示没有电脑，1表示有电脑，dp里的值就是和u联通的联通块值最小值是多少
用一个bool数组辅助记录一下这个dp状态可不可以被达到
转移的话就是树背包，\(dp[u][j][a] + dp[v][h][b] \rightarrow dp[u][j + h][a | b]\)
然后如果有父亲的话用\(dp[u][j][0]\)和\(dp[u][j][1]\)更新\(dp[u][j + 1][0]\)
没有的话就直接记录答案就行
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 5005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
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 * 2];
int head[MAXN],sumE;
bool vis[MAXN][MAXN][2],used[MAXN][2];
int64 dp[MAXN][MAXN][2],A[MAXN],g[MAXN][2];
int N,siz[MAXN],ans;
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void dfs(int u,int fa) {
if(A[u] < 0) {
vis[u][0][1] = 1;dp[u][0][1] = A[u];
}
if(A[u] > 0) {
vis[u][0][0] = 1;dp[u][0][0] = A[u];
}
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa) {
dfs(v,u);
}
}
memset(used,0,sizeof(used));
siz[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa) {
for(int j = 0 ; j <= siz[u] + siz[v] ; ++j) used[j][0] = used[j][1] = 0;
for(int j = 0 ; j <= siz[u] ; ++j) {
for(int a = 0 ; a <= 1 ; ++a) {
if(!vis[u][j][a]) continue;
for(int h = 0 ; h <= siz[v] ; ++h) {
for(int b = 0 ; b <= 1 ; ++b) {
if(!vis[v][h][b]) continue;
if(!used[j + h][a | b]) {
g[j + h][a | b] = dp[u][j][a] + dp[v][h][b];
used[j + h][a | b] = 1;
}
else g[j + h][a | b] = min(g[j + h][a | b],dp[u][j][a] + dp[v][h][b]);
}
}
}
}
for(int j = 0 ; j <= siz[u] + siz[v] ; ++j) {
for(int a = 0 ; a <= 1 ; ++a) {
vis[u][j][a] = used[j][a];
dp[u][j][a] = g[j][a];
}
}
siz[u] += siz[v];
}
}
if(fa) {
for(int j = siz[u] ; j >= 0 ; --j) {
if(vis[u][j][1] && dp[u][j][1] < 0) {dp[u][j + 1][0] = 0;vis[u][j + 1][0] = 1;}
if(vis[u][j][0]) {dp[u][j + 1][0] = 0;vis[u][j + 1][0] = 1;}
}
}
else {
for(int j = 0 ; j <= siz[u] ; ++j) {
if(vis[u][j][1] && dp[u][j][1] < 0) {ans = j;break;}
if(vis[u][j][0]) {ans = j;break;}
}
}
}
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(A[i]);
int u,v;
for(int i = 1 ; i < N ; ++i) {
read(u);read(v);
add(u,v);add(v,u);
}
dfs(1,0);
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}