Codeforces Round #267 (Div. 2)

QAQAQAQAQ

D题sb题没写出来（大雾）

QAQAQAQ

差点掉ratingQAQ

c题我能再wa多次吗，就打错个max的转移啊！QAQ

A.George and Accommodation

题意：给你a和b，问你a是否小于等于b-2

这。。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

int main() {
	int n=getint(), ans=0;
	while(n--) {
		int a=getint(), b=getint();
		if(b-2>=a) ++ans;
	}
	print(ans);
	return 0;
}

B.Fedor and New Game

题意：给你m+1个数让你判断所给的数的二进制形式与第m+1个数不想同的位数是否小于等于k，累计答案

能再水点吗。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=1005;
int a[N], my, ans;
int main() {
	int n=getint(), m=getint(), k=getint();
	for1(i, 1, m) read(a[i]);
	read(my);
	for1(i, 1, m) {
		int tot=0;
		for3(j, n-1, 0) {
			if(((1<<j)&a[i])!=((1<<j)&my)) ++tot;
		}
		if(tot<=k) ++ans;
	}
	print(ans);
	return 0;
}

C.George and Job

题意：给你n个数，让你分成k块不相交的大小为m的连续的块，然后求所有可行方案的最大值

设d[i, j]表示前i个数分成j块的最大值

d[i, j]=max{d[k, j-1]}+sum[i-m, i]，1<=k<=i-m

答案是max{d[i, k]}

这里是n^3的，我们考虑优化横n^2

设mx[i, j]表示max{d[k, j]} 1<=k<=j

然后转移变成

d[i, j]=mx[i-m, j-1]+sum[i-m, i]

而mx的转移是

mx[i, j]=max{mx[i-1, j], d[i, j]}

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%I64d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const ll getint() { ll r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const ll max(const ll &a, const ll &b) { return a>b?a:b; }
inline const ll min(const ll &a, const ll &b) { return a<b?a:b; }

const int N=5005;
int n, m, k;
ll sum[N], d[N], ans, mx[N][N], f[N], a[N];
int main() {
	read(n); read(m); read(k);
	for1(i, 1, n) read(a[i]), sum[i]=sum[i-1]+a[i];
	for1(i, m, n) d[i]=sum[i]-sum[i-m];
	for1(i, m, n) {
		for3(j, k, 1) {
			f[j]=mx[i-m][j-1]+d[i];
			mx[i][j]=max(mx[i-1][j], f[j]);
		}
		ans=max(ans, f[k]);
	}
	print(ans);
	return 0;
}

D.Fedor and Essay

字符串题，，，，

题意：给你一个文本串，不分大小，里边有n个单词，然后给你m个转换，求转换后的文本串的最小“r”的数量并且文本的长度最短

QAQ

有环啊啊 啊。。还要开long long

所以dfs之前我们要缩点。。。

然后缩点后我们dfs维护最小，然后输出答案。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=500005;
string tp;
map<string, int> mp;
char rdin[N];
int n, m, tot, d[N], ln[N], cnt, ihead[N], vis[N], mn[N], essay[N], FF[N], LL[N], scc, tm, top, q[N], len[N], belongs[N], U[N], V[N];
struct ED { int next, to; }e[N*4];
void add(int u, int v) {
	e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v;
}
int cal(char *s) {
	int len=strlen(s), ret=0;
	rep(j, len) if(s[j]=='r') ++ret;
	return ret;
}
int readin() {
	scanf("%s", rdin);
	int len=strlen(rdin);
	rep(j, len) if(rdin[j]>='A'&&rdin[j]<='Z') rdin[j]=rdin[j]-'A'+'a';
	tp=rdin;
	if(mp[tp]) return mp[tp];
	mp[tp]=++tot;
	ln[tot]=strlen(rdin);
	d[tot]=cal(rdin);
	return tot;
}
void dfs(int u) {
	vis[u]=1;
	int v;
	for(int i=ihead[u]; i; i=e[i].next) {
		if(!vis[v=e[i].to]) dfs(v);
		if(mn[v]<mn[u] || (mn[v]==mn[u] && len[v]<len[u])) {
			mn[u]=mn[v];
			len[u]=len[v];
		}
	}
}
void tarjan(int u) {
	LL[u]=FF[u]=++tm;
	vis[u]=1; q[++top]=u;
	int v;
	for(int i=ihead[u]; i; i=e[i].next) {
		v=e[i].to;
		if(!FF[v]) tarjan(v), LL[u]=min(LL[u], LL[v]);
		else if(vis[v]) LL[u]=min(LL[u], FF[v]);
	}
	if(FF[u]==LL[u]) {
		++scc;
		int x;
		do {
			x=q[top--];
			vis[x]=0;
			belongs[x]=scc;
			if(d[x]<mn[scc] || (d[x]==mn[scc]&&ln[x]<len[scc])) mn[scc]=d[x], len[scc]=ln[x];
		} while(x!=u);
	}
}
void rebuild() {
	CC(vis, 0); CC(ihead, 0);
	cnt=0;
	for1(i, 1, m)
		if(belongs[U[i]]!=belongs[V[i]]) add(belongs[U[i]], belongs[V[i]]);
}
int main() {
	read(n);
	for1(i, 1, n) essay[i]=readin();
	read(m);
	for1(i, 1, m) {
		int pos1=readin(), pos2=readin();
		add(pos1, pos2);
		U[i]=pos1, V[i]=pos2;
	}
	CC(mn, 0x3f); CC(len, 0x3f);
	for1(i, 1, tot) if(!FF[i]) tarjan(i);
	rebuild();
	for1(i, 1, scc) dfs(i);
	long long ans1=0, ans2=0;
	for1(i, 1, n) {
		int pos=belongs[essay[i]];
		ans1+=mn[pos];
		ans2+=len[pos];
	}
	printf("%I64d %I64d
", ans1, ans2);
	return 0;
}

E.没看题。。