【AtCoder】ARC092

C - 2D Plane 2N Points

把能连边的点找到然后跑二分图匹配即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
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) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int g[105][105];
int a[105],b[105],c[105],d[105],N,matk[105];
bool vis[105];
bool match(int u) {
    for(int i = 1 ; i <= 100 ; ++i) {
	if(g[u][i] && !vis[i]) {
	    vis[i] = 1;
	    if(!matk[i] || match(matk[i])) {
		matk[i] = u;
		return true;
	    }
	}
    }
    return false;
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);read(b[i]);}
    for(int i = 1 ; i <= N ; ++i) {read(c[i]);read(d[i]);}
    for(int i = 1 ; i <= N ; ++i) {
	for(int j = 1 ; j <= N ; ++j) {
	    if(a[i] < c[j] && b[i] < d[j]) g[i][j] = 1;
	}
    }
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
	memset(vis,0,sizeof(vis));
	if(match(i)) ++ans;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Two Sequences

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define MAXN 200005
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
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) {putchar('-');x = -x;}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int a[MAXN],b[MAXN];
int val[2][MAXN],len[2];
bool check(int p) {
    len[0] = len[1] = 0;
    for(int i = 1 ; i <= N ; ++i) {
        if(a[i] & (1 << p)) val[1][++len[1]] = a[i] & (1 << p) - 1;
        else val[0][++len[0]] = a[i] & (1 << p) - 1;
    }
    sort(val[1] + 1,val[1] + len[1] + 1);
    sort(val[0] + 1,val[0] + len[0] + 1);
    int64 res = 0;
    for(int i = 1 ; i <= N ; ++i) {
        int t = (1 << p) - (b[i] & (1 << p) - 1);
        int p0,p1;
        p0 = lower_bound(val[0] + 1,val[0] + len[0] + 1,t) - val[0] - 1;
        p1 = lower_bound(val[1] + 1,val[1] + len[1] + 1,t) - val[1] - 1;
        if(b[i] & (1 << p)) {res += p0;res += len[1] - p1;}
        else {res += p1;res += len[0] - p0;}
    }
    return res & 1;
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);}
    for(int i = 1 ; i <= N ; ++i) {read(b[i]);}
    int ans = 0;
    for(int i = 28 ; i >= 0 ; --i) {
        if(check(i)) ans |= (1 << i);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Both Sides Merger

显然是只能从\(f[i] = max(f[i - 2k] + a[i])\)转移过来

记录一下前一个从哪里转移过来，删一段区间的时候从中间开始删

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 1005
//#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;
int64 a[MAXN],dp[MAXN];
int fr[MAXN];
int op[MAXN],tot;

void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    int64 ans = a[1];
    for(int i = 1 ; i <= N ; ++i) {
	dp[i] = a[i];fr[i] = 0;
	for(int j = 2 ; j <= i ; j += 2) {
	    if(dp[i - j] + a[i] > dp[i]) {
		dp[i] = dp[i - j] + a[i];
		fr[i] = i - j;
	    }
	}
	ans = max(ans,dp[i]);
    }
    out(ans);enter;
    int a = N;
    for(int i = N ; i >= 1 ; --i) {
	if(dp[i] == ans) {
	    for(int j = N ; j > i ; --j) {op[++tot] = j;--a;}
	    int t = i,c = 0;
	    while(fr[t]) {
		++c;
		int mid = (fr[t] + 1 + a - c) >> 1;
		for(int j = mid ; j > fr[t] ; --j) {op[++tot] = j;}
		a -= (a - c - fr[t]);
		t = fr[t];
	    }
	    for(int j = 1 ; j < t ; ++j) op[++tot] = 1;
	    break;
	}
    }
    out(tot);enter;
    for(int i = 1 ; i <= tot ; ++i) {
	out(op[i]);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Two Faced Edges

mdzz，我不会卡常，走了

生气了