【AtCoder】ARC085

C - HSI

題解

\(E = 1900 * (N - M) + 100 * M + \frac{1}{2^{M}} E\)
\(E = 2^{M}(1900 * (N - M) + 100 * M)\)

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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);
}
int N,M;
void Solve() {
    read(N);read(M);
    int ans = M * 1900 + (N - M) * 100;
    for(int i = 1 ; i <= M ; ++i) ans *= 2;
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - ABS

簡單的minmax搜索，就是dp[i][0/1]表示前一張牌是第i張，0是X局面，1是Y局面
每次在X層取max，在Y層取min

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 2005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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);
}
int N,Z,W;
int a[MAXN],dp[MAXN][2];
int dfs(int on,int t) {
    if(dp[t][on] != -1) return dp[t][on];
    dp[t][on] = abs(a[t] - a[N]);
    for(int i = 1 ; i < N - t ; ++i) {
    if(on == 0) dp[t][on] = max(dp[t][on],dfs(on ^ 1,t + i));
    else dp[t][on] = min(dp[t][on],dfs(on ^ 1,t + i));
    }
    return dp[t][on];
}
void Solve() {
    read(N);read(Z);read(W);
    a[0] = W;
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    memset(dp,-1,sizeof(dp));
    out(dfs(0,0));
    enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - MUL

題解

至關於把邊權取反後求一個最大權閉合子圖

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 105
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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);
}
int N;
struct node {
    int to,next;
    int64 cap;
}E[MAXN * MAXN * 2];
int sumE = 1,head[MAXN],S,T;
int64 a[MAXN];
void add(int u,int v,int64 c) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    E[sumE].cap = c;
    head[u] = sumE;
}
void addtwo(int u,int v,int64 c) {
    add(u,v,c);add(v,u,0);
}
int gap[MAXN],dis[MAXN],last[MAXN];
int64 sap(int u,int64 aug) {
    if(u == T) return aug;
    int64 flow = 0;
    for(int i = last[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(E[i].cap > 0) {
        if(dis[v] + 1 == dis[u]) {
        int64 t = sap(v,min(E[i].cap,aug - flow));
        E[i].cap -= t;E[i ^ 1].cap += t;
        flow += t;
        if(flow == aug) return flow;
        if(dis[S] >= T) return flow;
        }
    }
    }
    if(--gap[dis[u]++] == 0) dis[S] = T;
    ++gap[dis[u]];
    last[u] = head[u];
    return flow;
}
void Solve() {
    read(N);
    int64 ans = 0,tmp = 0;
    for(int i = 1 ; i <= N ; ++i) {
    read(a[i]);
    ans += a[i];
    }
    S = N + 1,T = N + 2;
    for(int i = 1 ; i <= N ; ++i) {
    if(a[i] > 0) addtwo(i,T,a[i]);
    else if(a[i] < 0) {
        addtwo(S,i,-a[i]);
        tmp -= a[i];
    }
    }
    for(int i = 1 ; i <= N ; ++i) {
    int t = 2 * i;
    while(t <= N) {
        addtwo(i,t,2e9);
        t += i;
    }
    }
    for(int i = 1 ; i <= N + 2 ; ++i) last[i] = head[i];
    while(dis[S] < T) tmp -= sap(S,1e18);
    ans += tmp;
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - NRE

題解

原本想着外國人真是瞧不起中國的數據結構水平啊看我用數據結構艹過去這題

結果題解就是數據結構= =

就是把區間按右端點排序，而後每次掃到一個右端點用左端點取更新它

同時把這個區間的左端點\(dp[l - 1] - (l - 1) + sum[l - 1]\)的值更新給\([l,r]\)
一個點的dp值認爲是在不斷的dp時的dp值，還有是線段樹裏的最小值加上這個點的\(r - sum[r]\)

代碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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);
}
int N,Q;
int dp[MAXN],b[MAXN];
int sum[MAXN],cnt[MAXN];
vector<int> ed[MAXN];
struct node {
    int l,r,cov;
}tr[MAXN * 4];
void build(int u,int l,int r) {
    tr[u].l = l;tr[u].r = r;tr[u].cov = 2 * N;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(u << 1,l,mid);
    build(u << 1 | 1,mid + 1,r);
}
void Change(int u,int l,int r,int v) {
    if(tr[u].l == l && tr[u].r == r) {
    tr[u].cov = min(tr[u].cov,v);
    return;
    }
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(r <= mid) Change(u << 1,l,r,v);
    else if(l > mid) Change(u << 1 | 1,l,r,v);
    else {Change(u << 1,l,mid,v);Change(u << 1 | 1,mid + 1,r,v);}
}
int Query(int u,int pos,int v) {
    if(tr[u].l == tr[u].r) {
    v = min(v,tr[u].cov);
    return v + tr[u].l - sum[tr[u].l];
    }
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(pos <= mid) return Query(u << 1,pos,min(v,tr[u].cov));
    else return Query(u << 1 | 1,pos,min(v,tr[u].cov));
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
    read(b[i]);
    sum[i] = sum[i - 1] + b[i];
    }
    read(Q);
    int l,r;
    build(1,0,N);
    for(int i = 1 ; i <= Q ; ++i) {
    read(l);read(r);
    ed[r].pb(l);
    }
    dp[0] = 0;
    for(int i = 1 ; i <= N ; ++i) {
    dp[i] = i;
    if(b[i] == 0) dp[i] = min(dp[i - 1],dp[i]);
    else dp[i] = min(dp[i - 1] + 1,dp[i]);
    for(auto t : ed[i]) {
        dp[i] = min(min(Query(1,t - 1,2 * N),dp[t - 1]) + (i - t + 1) - (sum[i] - sum[t - 1]),dp[i]);
    }
    for(auto t : ed[i]) {
        int a = min(Query(1,t - 1,2 * N),dp[t - 1]);
        Change(1,t,i,a - (t - 1) + sum[t - 1]);
    }
    }
    out(dp[N]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}