【AtCoder】ARC073

時間 2019-12-09

標籤 AtCoder arc073 arc 简体版

原文原文鏈接

ARC 073

C - Sentou

直接線段覆蓋便可node

#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('\n')
#define eps 1e-10
#define MAXN 300005
//#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);
}
int N,T;
int ans;
void Solve() {
    read(N);read(T);
    int t,r;
    r = 0;
    for(int i = 1 ; i <= N ; ++i) {
        read(t);
        ans += t + T - max(t,r);
        r = t + T;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Simple Knapsack

把每一個揹包的體積改爲\(w_i - w_1\)c++

而後記錄\(dp[i][j]\)爲選了i個體積爲j的最大值優化

對於答案枚舉i而後取剩餘\(W - i \times w_1\)部分能取最多的便可ui

#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('\n')
#define eps 1e-10
#define MAXN 300005
//#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);
}
int N;
int64 dp[105][305],w[105],v[105],W;
void Solve() {
    read(N);read(W);
    for(int i = 1 ; i <= N ; ++i) {
        read(w[i]);read(v[i]);
    }
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = i ; j >= 1 ; --j) {
            for(int k = w[i] - w[1] ; k <= 3 * N ; ++k) {
                dp[j][k] = max(dp[j][k],dp[j - 1][k - (w[i] - w[1])] + v[i]);
            }
        }
    }
    int64 ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
        if(W < i * w[1]) break;
        ans = max(dp[i][min(W - i * w[1],1LL * 3 * N)],ans);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Ball Coloring

咱們認爲這2n個數最大爲\(MAX\)，最小爲\(MIN\)spa

發現\(R_{max}\)和\(B_{max}\)裏必定有一個\(MAX\)code

\(R_{min}\)和\(B_{min}\)裏必定有一個\(MIN\)排序

而後不失通常性，認爲有兩種狀況get

\(R_{max} = MAX,B_{min} = MIN\)it

這個時候咱們要\(B_{max}\)最小，\(R_{min}\)最大，咱們只須要兩個球裏選最大的塗紅，最小的塗藍便可class

\(R_{max} = MAX,R_{min} = MIN\)

咱們把全部袋中數值小的球塗成藍色，而後排序，逐個把藍色紅色互換，求每次藍球間的最小值

發現這兩種方式處理方式其實在第二種都包含了

#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('\n')
#define eps 1e-10
#define MAXN 400005
//#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);
}
int N;
pii p[MAXN];
multiset<int> sb,sr;
void Solve() {
    read(N);
    int x,y;
    for(int i = 1 ; i <= N ; ++i) {
        read(x);read(y);
        if(x > y) swap(x,y);
        p[i] = mp(x,y);
        sb.insert(x);sr.insert(y);
    }
    sort(p + 1,p + N + 1);
    int64 ans = 1e18;
    for(int i = 1 ; i <= N ; ++i) {
        int a = *(--sr.end()) - *sr.begin();
        int b = *(--sb.end()) - *sb.begin();
        ans = min(ans,1LL * a * b);
        sr.erase(sr.find(p[i].se));
        sr.insert(p[i].fi);
        sb.erase(sb.find(p[i].fi));
        sb.insert(p[i].se);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Many Moves

記錄狀態\(dp[x[i]][a]\)爲兩個塊分別在\(a\)和\(x[i]\)時最小的價值

咱們至關於對於\(dp[x[i]][x[i - 1]]\)特殊處理，只要求\(dp[x[i - 1]][1,x[i - 1] - 1]\)和\(dp[x[i - 1]][x[i - 1] +1,N]\)的後一維到\(x[i]\)的距離加dp值

而後其他位置都直接加上\(abs(x[i] - x[i - 1])\)

直接線段樹優化轉移

#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('\n')
#define eps 1e-10
#define MAXN 400005
//#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 {
    int64 val[2],lz;
    int L,R;
}tr[MAXN * 4];
int N,Q,A,B,x[MAXN];
void addlz(int u,int64 v) {
    tr[u].lz += v;
    for(int i = 0 ; i < 2 ; ++i) tr[u].val[i] += v;
}
void pushdown(int u) {
    if(tr[u].lz) {
        addlz(u << 1,tr[u].lz);
        addlz(u << 1 | 1,tr[u].lz);
        tr[u].lz = 0;
    }
}
void update(int u) {
    for(int i = 0 ; i < 2 ; ++i)
        tr[u].val[i] = min(tr[u << 1].val[i],tr[u << 1 | 1].val[i]);
}
void build(int u,int l,int r) {
    tr[u].L = l;tr[u].R = r;
    if(l == r) {
        tr[u].val[0] = 1e16 - l;
        tr[u].val[1] = 1e16 + l;
        return;
    }
    int mid = (l + r) >> 1;
    build(u << 1,l,mid);
    build(u << 1 | 1,mid + 1,r);
    update(u);
}
void Change_pos(int u,int p,int64 v) {
    if(tr[u].L == tr[u].R) {
        tr[u].val[0] = v - tr[u].L;
        tr[u].val[1] = v + tr[u].L;
        return;
    }
    pushdown(u);
    int mid = (tr[u].L + tr[u].R) >> 1;
    if(p <= mid) Change_pos(u << 1,p,v);
    else Change_pos(u << 1 | 1,p,v);
    update(u);
}
void Add_Range(int u,int l,int r,int64 v) {
    if(l > r) return;
    if(tr[u].L == l && tr[u].R == r) {
        addlz(u,v);return;
    }
    pushdown(u);
    int mid = (tr[u].L + tr[u].R) >> 1;
    if(r <= mid) Add_Range(u << 1,l,r,v);
    else if(l > mid) Add_Range(u << 1 | 1,l,r,v);
    else {Add_Range(u << 1,l,mid,v);Add_Range(u << 1 | 1,mid + 1,r,v);}
    update(u);
}
int64 Query(int u,int l,int r,int id) {
    if(l > r) return 1e16;
    if(tr[u].L == l && tr[u].R == r) return tr[u].val[id];
    pushdown(u);
    int mid = (tr[u].L + tr[u].R) >> 1;
    if(r <= mid) return Query(u << 1,l,r,id);
    else if(l > mid) return Query(u << 1 | 1,l,r,id);
    else {return min(Query(u << 1,l,mid,id),Query(u << 1 | 1,mid + 1,r,id));}
}
int64 Getans(int u) {

    if(tr[u].L == tr[u].R) return tr[u].val[0] + tr[u].L;
    pushdown(u);
    return min(Getans(u << 1),Getans(u << 1 | 1));
}
void Solve() {
    read(N);read(Q);read(A);read(B);
    build(1,1,N);
    x[0] = A;
    Change_pos(1,B,0);
    for(int i = 1 ; i <= Q ; ++i) {
        read(x[i]);
        int64 trans = 1e18;
        trans = min(x[i] + Query(1,1,x[i],0),trans);
        trans = min(Query(1,x[i] + 1,N,1) - x[i],trans);
        Change_pos(1,x[i - 1],trans);
        Add_Range(1,1,x[i - 1] - 1,abs(x[i] - x[i - 1]));
        Add_Range(1,x[i - 1] + 1,N,abs(x[i] - x[i - 1]));
    }
    out(Getans(1));enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

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