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
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;
}
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;
}
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;
}
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\n", ans1, ans2);
return 0;
}
E.没看题。。