ARC 074
C - Chocolate Bar
直接枚举第一刀横切竖切，然后另一块要求如果横切分成(H / 2)竖切分成(W/2)即可
#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('
')
#define eps 1e-10
#define MAXN 200005
//#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 H,W,ans = 1e18;
int64 maxdec(int64 a,int64 b,int64 c) {
return max(max(abs(a - b),abs(a - c)),abs(b - c));
}
void Solve() {
read(H);read(W);
for(int i = 1 ; i < H ; ++i) {
int64 h = (H - i) / 2;
ans = min(ans,maxdec(i * W,h * W,(H - i - h) * W));
int64 t = W / 2;
ans = min(ans,maxdec(i * W,t * (H - i),(W - t) * (H - i)));
}
for(int i = 1 ; i < W ; ++i) {
int64 h = (W - i) / 2;
ans = min(ans,maxdec(i * H,h * H,(W - i - h) * H));
int64 t = H / 2;
ans = min(ans,maxdec(i * H,t * (W - i),(H - t) * (W - i)));
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
D - 3N Numbers
直接算前i个数最大的N个是多少，后i个数最小的是多少，可以用set
#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('
')
#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);
}
struct cmp {
bool operator () (const int &a,const int &b) const {
return a > b;
}
};
int N,a[MAXN];
int64 f[MAXN],b[MAXN];
multiset<int> s2;
multiset<int,cmp> s1;
void Solve() {
read(N);
for(int i = 1 ; i <= 3 * N ; ++i) read(a[i]);
int64 sum = 0;
for(int i = 1 ; i <= 3 * N ; ++i) {
s1.insert(a[i]);sum += a[i];
if(s1.size() > N) {
auto t = *(--s1.end());
s1.erase(--s1.end());
sum -= t;
}
f[i] = sum;
}
sum = 0;
for(int i = 3 * N ; i >= 1 ; --i) {
s2.insert(a[i]);sum += a[i];
if(s2.size() > N) {
auto t = *(--s2.end());
s2.erase(--s2.end());
sum -= t;
}
b[i] = sum;
}
int64 ans = f[N] - b[N + 1];
for(int i = N + 1 ; i <= 2 * N ; ++i) {
ans = max(ans,f[i] - b[i + 1]);
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - RGB Sequence
就是记录(dp[r][g][b])为上一次出现r的位置，出现g的位置和出现b的位置，把不合法的状态标成0
#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('
')
#define eps 1e-10
//#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 N,M;
int dp[305][305][305],sum[3][305][305];
vector<pii > v[305];
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;
}
bool check(int r,int g,int b,int len) {
for(int i = 0 ; i < v[len].size() ; ++i) {
pii t = v[len][i];
int cnt = 0;
if(r >= t.fi) ++cnt;
if(g >= t.fi) ++cnt;
if(b >= t.fi) ++cnt;
if(cnt != t.se) return false;
}
return true;
}
void increase(int r,int g,int b) {
sum[0][g][b] = inc(sum[0][g][b],dp[r][g][b]);
sum[1][r][b] = inc(sum[1][r][b],dp[r][g][b]);
sum[2][r][g] = inc(sum[2][r][g],dp[r][g][b]);
}
void Solve() {
read(N);read(M);
int l,r,x;
for(int i = 1 ; i <= M ; ++i) {
read(l);read(r);read(x);
v[r].pb(mp(l,x));
}
dp[0][0][0] = 1;
sum[0][0][0] = 1;
sum[1][0][0] = 1;
sum[2][0][0] = 1;
for(int i = 1 ; i <= N ; ++i) {
for(int j = 0 ; j < i ; ++j) {
for(int k = 0 ; k < i ; ++k) {
dp[i][j][k] = sum[0][j][k];
dp[j][i][k] = sum[1][j][k];
dp[j][k][i] = sum[2][j][k];
if(!check(i,j,k,i)) dp[i][j][k] = 0;
if(!check(j,i,k,i)) dp[j][i][k] = 0;
if(!check(j,k,i,i)) dp[j][k][i] = 0;
}
}
memset(sum,0,sizeof(sum));
for(int j = 0 ; j < i ; ++j) {
for(int k = 0 ; k < i ; ++k) {
increase(i,j,k);
increase(j,i,k);
increase(j,k,i);
}
}
}
int ans = 0;
for(int i = 0 ; i < N ; ++i) {
for(int j = 0 ; j < N ; ++j) {
ans = inc(ans,dp[N][i][j]);
ans = inc(ans,dp[i][N][j]);
ans = inc(ans,dp[i][j][N]);
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
F - Lotus Leaves
行列拆点，对于然后每个叶子新建一个点往对应的行列连边
对于起点和重点的边建成无穷大
然后跑网络流求最小割即可
#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('
')
#define eps 1e-10
#define MAXN 200005
//#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 {
int to,next,cap;
}E[MAXN];
int sumE = 1,head[MAXN];
int H,W,Ncnt,S,T;
char s[105][105];
void add(int u,int v,int c) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].cap = c;
head[u] = sumE;
}
void addtwo(int u,int v,int c) {
add(u,v,c);add(v,u,0);
}
int gap[MAXN],dis[MAXN];
int isap(int u,int aug) {
if(u == T) return aug;
int flow = 0;
for(int i = head[u] ; i ;i = E[i].next) {
int v = E[i].to;
if(E[i].cap) {
if(dis[v] + 1 == dis[u]) {
int t = isap(v,min(aug - flow,E[i].cap));
flow += t;
E[i].cap -= t;
E[i ^ 1].cap += t;
if(aug == flow) return flow;//这块写错了，如果能满流的话应该返回而不是让dis++
if(dis[S] >= Ncnt) return flow;
}
}
}
--gap[dis[u]];
if(!gap[dis[u]]) {dis[S] = Ncnt;return flow;}
dis[u]++;
++gap[dis[u]];
return flow;
}
void Solve() {
read(H);read(W);
for(int i = 1 ; i <= H ; ++i) {
scanf("%s",s[i] + 1);
}
Ncnt = H + W;
pii p[2];
for(int i = 1 ; i <= H ; ++i) {
for(int j = 1 ; j <= W ; ++j) {
if(s[i][j] != '.') {
++Ncnt;
int t = 0;
if(s[i][j] == 'o') t = 1;
else t = 0x7fffffff;
addtwo(Ncnt,i,t);addtwo(i,Ncnt,t);
addtwo(j + H,Ncnt,t);addtwo(Ncnt,j + H,t);
if(s[i][j] == 'S') {p[0] = mp(i,j);S = Ncnt;}
if(s[i][j] == 'T') {p[1] = mp(i,j);T = Ncnt;}
}
}
}
if(p[0].fi == p[1].fi || p[0].se == p[1].se) {puts("-1");return;}
int ans = 0;
while(dis[S] < Ncnt) ans += isap(S,0x7fffffff);
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}