gym101908 [2018-2019 ACM-ICPC Brazil Subregional Programming Contest] 题解
感觉难度和邀请赛类似,题目质量也低于国内邀请赛,(题面/数据不出锅的情况下)
https://codeforces.com/gym/101908
A.大概是莫比乌斯之类的,不会
B:博弈,不会
C:欧拉公式+二维偏序
首先,根据平面图欧拉公式,可推导出答案为$n+m+1+$交叉的数量
交叉的数量由三部分构成,横竖交叉数,横横交叉数和竖竖交叉数
显然,前者为$nm$,后两个为简单的二维偏序问题,
比较扯淡的是,先T了一发,从map改完二分,交的时候才意识到是没关同步
#include<bits/stdc++.h>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define fi first
#define se second
#define pii pair<int,int>
#define all(x) x.begin(),x.end()
using namespace std;//head
const int maxn=1e6+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
ll x,y;
class bit{public:
int node[maxn];
inline int lb(int x) {return x&(-x);}
void init(int n){fill_n(node,n+1,0);}
inline void update(int pos,int val,int n){for(;pos>0&&pos<=n;pos+=lb(pos)) node[pos]+=val;}
inline int ask(int pos,int sum,int n){for(;pos>0;pos-=lb(pos)) sum+=node[pos];return sum;}
inline int query(int l,int r,int n){return ask(r,0,n)-ask(l-1,0,n);}
}tree;
int px[maxn];
ll solve(int n){
int cntx=0;
vector<pii>h;
rep(i,1,n){
int a,b;cin>>a>>b;
h.emplace_back(a,b);
px[++cntx]=b;
}
sort(px+1,px+1+cntx);
cntx=unique(px+1,px+1+cntx)-px-1;
for(auto &i:h){
i.se=lower_bound(px+1,px+1+cntx,i.se)-px;
}
ll cnt=0;
sort(all(h));
tree.init(cntx);
for(auto &i:h){
cnt+=tree.query(i.se+1,cntx,cntx);
tree.update(i.se,1,cntx);
}
return cnt;
}
int main() {IO;
cin>>x>>y>>n>>m;
ll ans=solve(n)+solve(m)+n+m+1+1ll*n*m;
cout<<ans<<endl;
}
D题:签到
E题:签到(30ms跑1e8已经是cf的基本操作了吗?)
F题:状压DP
先对于节目,按结束时间排序,然后依次枚举:节目数量,节目,选择的$Festival$状态,
复杂度$(2^n)(\sum{M_i})^2$,大概是1e9,最后运行时间500ms,只能说CF评测机牛逼了
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 86500;
int dp[1100][1<<10];
struct node {int st, et, val, id;}p[1100];
bool cmp(node x, node y) {return x.et < y.et;}
int main() {
std::ios::sync_with_stdio(false);
int n,cnt = 0, ma = 0,ans = -1;
cin >> n;
for(int i = 0; i < n; i++) {
int k;cin >> k;
for(int j = 0; j < k; j++) {
int a, b, c, d;
cin >> a >> b >> c; d = i;
p[++cnt] = {a, b, c, d};
ma = max(ma, b);
}
}
sort(p+1, p+1+cnt, cmp);
memset(dp, -1, sizeof dp);
dp[0][0] = 0;
for(int i = 1; i <= cnt; i++) {
dp[i][0] = 0;
for(int j = 0; j < i; j++) {
if(p[i].st < p[j].et) continue;
for(int k = 1; k < (1<<n); k++) {
if((k&(1<<p[i].id))) {
if(dp[j][k] != -1) dp[i][k] = max(dp[i][k], dp[j][k]+p[i].val);
if(dp[j][k-(1<<p[i].id)] != -1) dp[i][k] = max(dp[i][k], dp[j][k-(1<<p[i].id)]+p[i].val);
}
}
}
ans = max(ans, dp[i][(1<<n)-1]);
}
cout << ans << "\n";
}
经过研究,可以优化为$(2^n)\sum{M_i}$,实测31ms
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 86500;
int dp[1100][1<<10], ma[1100][1<<10],n, cnt;
struct node {int st, et, val, id;}p[1100];
bool cmp(node x, node y) {return x.et < y.et;}
int f(int R, int k) {
int ans = 0, l = 1, r = R-1;
while(l <= r) {
int mid = (l+r)>>1;
if(p[mid].et <= k) ans = mid, l = mid+1;
else r = mid-1;
}
return ans;
}
int main() {
std::ios::sync_with_stdio(false);
cin >> n;
for(int i = 0; i < n; i++) {
int k;cin >> k;
for(int j = 0; j < k; j++) {
int a, b, c, d;
cin >> a >> b >> c; d = i;
p[++cnt] = {a, b, c, d};
}
}
sort(p+1, p+1+cnt, cmp);
memset(dp, -1, sizeof dp);
memset(ma, -1, sizeof ma);
ma[0][0] = dp[0][0] = 0;
int ans = -1;
for(int i = 1; i <= cnt; i++) {
dp[i][0] = 0;
int sign = f(i, p[i].st);
for(int k = 1; k < (1<<n); k++) {
if(!(k&(1<<p[i].id))) continue;
if(ma[sign][k] != -1) dp[i][k] = max(dp[i][k], ma[sign][k]+p[i].val);
if(ma[sign][k-(1<<p[i].id)] != -1) dp[i][k] = max(dp[i][k], ma[sign][k-(1<<p[i].id)]+p[i].val);
}
for(int k = 0; k < (1<<n); k++) ma[i][k] = max(ma[i-1][k], dp[i][k]);
}
cout << ma[cnt][(1<<n)-1] << "\n";
return 0;
}
G:二分+最小割
裸题,显然答案为使用过的最大的边,那么二分答案,看最小割是否等于$\sum{D_i}$即可,数组开小RE了一发
#include<bits/stdc++.h>
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
const int INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k,root;
class mxf{public:
struct node{int to,next;int cap;}e[maxm<<1];
int cur[maxn],head[maxn],que[maxn],dis[maxn],nume=1,s,t;
inline void adde(int a,int b,int c){e[++nume]={b,head[a],c};head[a]=nume;}
inline void add(int a,int b,int c){adde(a,b,c);adde(b,a,0);}
void init(int n=maxn-1){memset(head,0,(n+1)<<2);nume=1;}
bool bfs(){
memset(dis,-1,(t+1)<<2);
dis[t]=0,que[0]=t;
int tp=0,ed=1;
while(tp!=ed){
int now=que[tp++];tp%=maxn;
for(int i=head[now];i;i=e[i].next){
int to=e[i].to;
if(dis[to]==-1&&e[i^1].cap){
dis[to]=dis[now]+1;
if(to==s) return true;
que[ed++]=to;ed%=maxn;
}
}
}
return false;
}
int dfs(int now,int flow=0x3f3f3f3f){
if(now==t||flow==0) return flow;
int use=0;
for(int &i=head[now];i&&use!=flow;i=e[i].next){
int to=e[i].to;
if(dis[to]+1!=dis[now])continue;
int tmp=dfs(to,min(e[i].cap,flow-use));
e[i].cap-=tmp,e[i^1].cap+=tmp,use+=tmp;
}
if(!use) dis[now]=-1;
return use;
}
int getflow(int ss,int tt){
s=ss,t=tt;int ans=0;
memcpy(cur,head,(t+1)<<2);
while(bfs()){
ans+=dfs(s);
memcpy(head,cur,(t+1)<<2);
}
return ans;
}
}net;
int need[maxn],have[maxn];
int sum,a[maxn],b[maxn],c[maxn];
bool check(int mid){
net.init(n+m+10);
int ss=n+m+1,tt=ss+1;
rep(i,1,k)if(c[i]<=mid)net.add(a[i]+n,b[i],INF);
rep(i,1,n)net.add(i,tt,need[i]);
rep(i,1,m)net.add(ss,i+n,have[i]);
return net.getflow(ss,tt)>=sum;
}
int main() {IO;
cin>>n>>m>>k;
rep(i,1,n)(cin>>need[i]),sum+=need[i];
rep(i,1,m)cin>>have[i];
rep(i,1,k)cin>>b[i]>>a[i]>>c[i];
int l=1,r=1e6;
int ans=-1;
while(l<=r){
int mid=(l+r)>>1;
if(check(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
cout<<ans<<endl;
}
H题:没看
I题:签到
J题:没看
I题:计算几何,不会
L题:树上链交裸题
一种做法是用树链剖分,单次查询是$log^2n$,很裸也很简单
但是这次我尝试写了单次$logn$直接讨论情况的做法,由于本题$Q$很小,拉不开差距,
具体情况大概分为3类,讨论了很多发才过
#include<bits/stdc++.h>
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define forn(i,x) for(int i=head[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;//head
const int maxn=2e5+10,maxm=2e6+10;
int casn,n,m,k,root;
class graph{public:
struct node{int to,next;}e[maxn<<1];
int head[maxn],nume,dfn[maxn];
inline void add(int a,int b){
e[++nume]={b,head[a]};
head[a]=nume;
}
int ltop[maxn],fa[maxn],deep[maxn];
int sz[maxn],remp[maxn];
int son[maxn],cnt;
void init(int n){rep(i,1,n) head[i]=0;cnt=0,nume=1;}
void dfs1(int now=root,int pre=root,int d=0){
deep[now]=d,fa[now]=pre,sz[now]=1,son[now]=0;
forn(i,now){
int to=e[i].to;
if(to!=pre) {
dfs1(to,now,d+1);
sz[now]+=sz[to];
if(sz[to]>sz[son[now]]) son[now]=to;
}
}
}
void dfs2(int now=root,int pre=root,int sp=root){
ltop[now]=sp;dfn[now]=++cnt;remp[cnt]=now;
if(son[now]) dfs2(son[now],now,sp);
forn(i,now){
int to=e[i].to;
if(to!=son[now]&&to!=pre) dfs2(to,now,to);
}
}
void getchain(){dfs1();dfs2();}
int lca(int x,int y){
for(;ltop[x]!=ltop[y];deep[ltop[x]]>deep[ltop[y]]?x=fa[ltop[x]]:y=fa[ltop[y]]);
return deep[x]<deep[y]?x:y;
}
inline int getdis(int a,int b){return deep[a]+deep[b]-2*deep[lca(a,b)];}
bool check(int a,int b){return dfn[a]<=dfn[b]&&dfn[a]+sz[a]-1>=dfn[b]+sz[b]-1;}
int getans(int a1,int a2,int b1,int b2){
int ra=lca(a1,a2);
bool f1=check(ra,b1),f2=check(ra,b2);
if(!f1&&!f2) return 0;
if(f1&&f2){
int rb=lca(b1,b2);
if(!( check(rb,a1)||check(rb,a2)))return 0;
int r1=lca(a1,b1),r2=lca(a1,b2),r3=lca(a2,b1),r4=lca(a2,b2);
if(r1==r3&&r2==r4) return 1;
return getdis(r1==ra?r3:r1,r2==ra?r4:r2)+1;
}
if(!f1)swap(b1,b2);
int r1=lca(a1,b1),r3=lca(a2,b1);
return getdis(r1==ra?r3:r1,ra)+1;
}
}g;
int d[maxn],a,b,a1,a2,b1,b2;
int main() {IO;
cin>>n>>m;
rep(i,2,n){
cin>>a>>b;
g.add(a,b);g.add(b,a);
d[a]++,d[b]++;
}
rep(i,1,n) if(d[i]>1) {root=i;break;}
g.getchain();
while(m--){
cin>>a1>>a2>>b1>>b2;
int ans=g.getans(a1,a2,b1,b2);
cout<<ans<<endl;
}
}
M题:没看
