【AtCoder】ARC083

C - Sugar Water

计算一下可以达到水是多少,可以到达的糖是多少
枚举水,然后加最多能加的糖,是\(min(F - i *100,E * 100)\),计算密度,和前一个比较就行

#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 MAXN 100005
#define eps 1e-12
//#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 A,B,C,D,E,F;
int w[35],t[3005];
void Solve() {
    read(A);read(B);read(C);read(D);read(E);read(F);
    w[0] = 1;
    for(int i = 1 ; i <= F / 100 ;  ++i) {
        if(i >= A) w[i] |= w[i - A];
        if(i >= B) w[i] |= w[i - B];
    }
    t[0] = 1;
    for(int i = 1 ; i <= F ; ++i) {
        if(i >= C) t[i] |= t[i - C];
        if(i >= D) t[i] |= t[i - D];
    }
    t[0] = 0;
    for(int i = 1 ; i <= F ; ++i) {
        if(t[i]) t[i] = i;
        else t[i] = t[i - 1];
    }
    int a = A * 100,b = 0;
    for(int i = 1 ; i <= F / 100 ; ++i) {
        if(!w[i]) continue;
        int rem = min(F - i * 100,i * E);
        int s = t[rem],p = i * 100 + s;
        if(s * a > b * p) {a = p;b = s;}
    }
    out(a);space;out(b);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Restoring Road Network

很容易发现边肯定都被涵盖在最短路的数值里(如果要构造出一个价值最小的图)

我们从小到大加边,对于一个(u,v)最短路,如果这个最短路不用额外边,那么就是用一个别的点k,(u,k) + (k,v) = (u,v)

如果算出来的最短路小于给定值,那么就不合法

#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 MAXN 100005
#define eps 1e-12
//#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);
}
int64 A[305][305];
int64 B[305][305],ans;
pii p[100005];
int tot,N;
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = 1 ; j <= N ; ++j) {
            read(A[i][j]);
            if(i != j) B[i][j] = 1e16;
            if(i < j) p[++tot] = mp(i,j);
        }
    }
    sort(p + 1,p + tot + 1,[](pii a,pii b){return A[a.fi][a.se] < A[b.fi][b.se];});
    for(int i = 1 ; i <= tot ; ++i) {
        int u = p[i].fi,v = p[i].se;
        for(int j = 1 ; j <= N ; ++j) {
            B[u][v] = min(B[u][j] + B[j][v],B[u][v]);
        }
        B[v][u] = B[u][v];
        if(B[u][v] < A[u][v]) {puts("-1");return;}
        if(B[u][v] == A[u][v]) continue;
        ans += A[u][v];
        B[u][v] = B[v][u] = A[u][v];
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Bichrome Tree

我们希望除了和该点同色的值之外,另一种颜色的值总和尽可能的小,这样容易达成目标
可以直接做一个背包dp就行,每次尽量取最小的值当做另一种颜色的值

#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 MAXN 5005
#define eps 1e-12
//#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,P[MAXN],X[MAXN];
int dp[MAXN],sum[MAXN];
bool f[MAXN][MAXN];
void Solve() {
    read(N);
    for(int i = 2 ; i <= N ; ++i) read(P[i]);
    for(int i = 1 ; i <= N ; ++i) read(X[i]);
    for(int i = 1 ; i <= N ; ++i) f[i][0] = 1;
    for(int i = N ; i >= 1 ; --i) {
        bool flag = 0;
        for(int j = X[i] ; j >= 0 ; --j) {
            if(f[i][j]) {
                dp[i] = sum[i] - j;
                flag = 1;
                break;
            }
        }
        if(!flag) {
            puts("IMPOSSIBLE");return;
        }
        if(P[i]) {
            for(int j = X[P[i]] ; j >= 0 ; --j) {
                bool t = 0;
                if(j >= X[i]) t |= f[P[i]][j - X[i]];
                if(j >= dp[i]) t |= f[P[i]][j - dp[i]];
                f[P[i]][j] = t;
            }
            sum[P[i]] += X[i] + dp[i];
        }
    }
    puts("POSSIBLE");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Collecting Balls

行列拆点,有一个点就行列之间连边

连出的图肯定是很多个基环外向树森林

相当于给每个点分配一个边,给环上的点分配边就两种方式,每种分别计算方案数

分配完之后每个点假如横向,这个点往横向之前所有的点连一条边
竖向同理

然后就是求一个树的dfs序个数

最后把所有联通块的答案合出来就行

#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 MAXN 200005
#define eps 1e-12
//#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 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;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
void update(int &x,int y) {
    x = inc(x,y);
}
struct node {
    int to,next;
}E[MAXN * 4];
int x[MAXN],y[MAXN],N,fac[MAXN],invfac[MAXN];
map<pii,int> zz;
int head[MAXN],sumE;
bool vis[MAXN];
int sum,Ncnt,fa[MAXN],s,t,pos[MAXN],par[MAXN];
vector<int> c[MAXN],r[MAXN],rec,cyc;
int rpos[MAXN],cpos[MAXN],siz[MAXN];
vector<int> son[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
pii depos(int a,int b) {
    if(a > b) swap(a,b);
    b -= N;
    return mp(a,b);
}
bool dfs(int u) {
    ++Ncnt;
    bool flag = 0;
    vis[u] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
        ++sum;
        int v = E[i].to;
        if(!vis[v]) {
            fa[v] = u;
            if(dfs(v)) {
                flag = 1;
                pos[zz[depos(u,v)]] = u;
            }
        }
        else if(v != fa[u] && !flag){
            s = u,t = v;
            flag = 1;
            rec.pb(zz[depos(u,v)]);
        }
    }
    if(!flag && fa[u]) {
        pos[zz[depos(u,fa[u])]] = u;
    }
    if(fa[u]) rec.pb(zz[depos(u,fa[u])]);
    return flag;
}
int Calc(int u) {
    siz[u] = 1;
    int res = 1;
    for(auto v : son[u]) {
        res = mul(res,Calc(v));
        res = mul(res,invfac[siz[v]]);
        siz[u] += siz[v];
    }
    res = mul(res,fac[siz[u] - 1]);
    return res;
}
int Process() {
    int res = 0;
    for(auto t : rec) par[t] = 0;
    for(auto t : rec) {
        if(pos[t] > N) {
            for(int i = 0 ; i < cpos[t] ; ++i) {
                par[zz[mp(c[y[t]][i],y[t])]] = t;
            }
        }
        else {
            for(int i = 0 ; i < rpos[t] ; ++i) {
                par[zz[mp(x[t],r[x[t]][i])]] = t;
            }
        }
        son[t].clear();
    }
    son[0].clear();
    for(auto t : rec) {
        son[par[t]].pb(t);
    }
    return Calc(0);
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= 2 * N ; ++i) {
        read(x[i]);read(y[i]);add(x[i],y[i] + N);add(y[i] + N,x[i]);
        zz[mp(x[i],y[i])] = i;
        r[x[i]].pb(y[i]);c[y[i]].pb(x[i]);
    }
    for(int i = 1 ; i <= N ; ++i) {
        sort(r[i].begin(),r[i].end());
        sort(c[i].begin(),c[i].end());
        for(int j = 1 ; j < r[i].size() ; ++j) rpos[zz[mp(i,r[i][j])]] = j;
        for(int j = 1 ; j < c[i].size() ; ++j) cpos[zz[mp(c[i][j],i)]] = j;
    }
    fac[0] = 1;
    for(int i = 1 ; i <= 2 * N ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[2 * N] = fpow(fac[2 * N],MOD - 2);
    for(int i = 2 * N - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    int ans = fac[2 * N];
    for(int i = 1 ; i <= 2 * N ; ++i) {
        if(!vis[i]) {
            sum = 0;Ncnt = 0;
            rec.clear();
            dfs(i);
            ans = mul(ans,invfac[Ncnt]);
            if(Ncnt * 2 != sum) {puts("0");return;}
            int p = s;
            cyc.clear();
            while(1) {
                cyc.pb(p);
                if(p == t) break;
                p = fa[p];
            }
            int tmp = 0;
            cyc.pb(s);
            for(int i = 0 ; i < cyc.size() - 1 ; ++i) {
                pos[zz[depos(cyc[i],cyc[i + 1])]] = cyc[i];
            }
            update(tmp,Process());
            for(int i = cyc.size() - 1 ; i >= 1 ; --i) {
                pos[zz[depos(cyc[i],cyc[i - 1])]] = cyc[i];
            }
            update(tmp,Process());
            ans = mul(ans,tmp);
        }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

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转载自www.cnblogs.com/ivorysi/p/10374088.html
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