B. Connecting Universities
大意: 给定树, 给定2*k个点, 求将2*k个点两两匹配, 每个匹配的贡献为两点的距离, 求贡献最大值
单独考虑每条边$(u,v)$的贡献即可, 最大贡献显然是左右两侧点的最小值.
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head const int N = 1e6+10;
int n, k;
int a[N], sz[N];
vector<int> g[N];
ll ans; void dfs(int x, int fa) {
sz[x] = a[x];
for (int y:g[x]) if (y!=fa) {
dfs(y,x), sz[x]+=sz[y];
ans += min(sz[y], k-sz[y]);
}
} int main() {
scanf("%d%d", &n, &k),k*=2;
REP(i,1,k) {
int t;
scanf("%d", &t);
a[t] = 1;
}
REP(i,2,n) {
int u, v;
scanf("%d%d", &u, &v);
g[u].pb(v),g[v].pb(u);
}
dfs(1,0);
printf("%lld\n", ans);
}
C. Break Up
大意: 无向有权图有重边自环, 求删除两条边使得s与t不连通, 且两条边的边权和最小.
先求出任意一条最短路径, 边数显然不超过$n$, 暴力枚举这$n$条边然后再tarjan即可, 复杂度O(n(m+n))
算是挺简单的了, 还是打了好久, 一直卡在怎么判断删除一条边后是否连通, 后来发现tarjan后从s->t经过的桥一定是一条链, 所以直接dfs就好了, 最后还要注意边权1e9+1e9爆掉0x3f3f3f3f了.
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = ~0u>>1;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head const int N = 3e4+10;
int n, m, S, T;
int w[N];
struct _ {int to,id;} fa[N];
vector<_> g[N];
int dfn[N], low[N], isbridge[N], clk;
void tarjan(int x, int fa, int z) {
dfn[x]=low[x]=++clk;
for (auto &&e:g[x]) if (e.id!=z) {
int y = e.to, id = e.id;
if (!dfn[y]) {
tarjan(y,id,z);
low[x]=min(low[x],low[y]);
if (low[e.to]>dfn[x]) isbridge[id]=1;
} else if (dfn[y]<dfn[x]&&id!=fa) {
low[x]=min(low[x],dfn[y]);
}
}
}
int vis[N], c[N];
int dfs(int x) {
if (x==T) return 1;
for (auto e:g[x]) if (!vis[e.id]) {
vis[e.id] = 1;
if (dfs(e.to)) return c[e.id] = 1;
}
return 0;
} int main() {
scanf("%d%d%d%d", &n, &m, &S, &T);
REP(i,1,m) {
int u, v;
scanf("%d%d%d", &u, &v, w+i);
g[u].pb({v,i}), g[v].pb({u,i});
}
queue<int> q;
fa[S].to=-1, q.push(S);
while (q.size()) {
int x = q.front(); q.pop();
for (auto &&e:g[x]) if (!fa[e.to].to) {
fa[e.to]={x,e.id}, q.push(e.to);
}
}
if (!fa[T].to) return puts("0\n0"),0;
int ans = INF;
vector<int> vec;
for (int x=T; x!=S; x=fa[x].to) {
int id = fa[x].id;
memset(vis,0,sizeof vis);
memset(c,0,sizeof c);
vis[id] = 1;
if (!dfs(S)) {
if (ans>w[id]) ans = w[id],vec.clear(),vec.pb(id);
continue;
}
memset(dfn,0,sizeof dfn);
memset(isbridge,0,sizeof isbridge);
clk = 0;
tarjan(S,0,id);
REP(i,1,m) if (c[i]&&isbridge[i]&&ans>w[id]+w[i]) {
ans=w[id]+w[i];
vec.clear();
vec.pb(id), vec.pb(i);
}
}
if (ans==INF) return puts("-1"),0;
printf("%d\n%d\n", ans, int(vec.size()));
for (int t:vec) printf("%d ", t); hr;
}
D. Huffman Coding on Segment
莫队一下, 然后将出现次数小于等于$\sqrt{n}$的暴力合, 其余的用堆合, 复杂度$O(m\sqrt{n}logn)$, 看了下最优解, 好像可以排序一下省去堆从而优化掉一个log
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head #ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif int n, m, sqn;
int blo[N], cnt[N], sum[N], s[N], a[N];
struct _ {
int l,r,id;
bool operator < (const _ & rhs) const {
return blo[l]^blo[rhs.l]?l<rhs.l:blo[l]&1?r<rhs.r:r>rhs.r;
}
} e[N];
ll ans[N];
vector<int> q; void upd(int x, int d) {
--sum[cnt[x]];
cnt[x]+=d;
++sum[cnt[x]];
} ll calc() {
ll ans = 0;
REP(i,1,sqn) s[i] = sum[i];
priority_queue<int,vector<int>,greater<int> > Q;
int pre = 0;
REP(i,1,sqn) if (s[i]) {
if (pre) {
int x = pre+i;
ans += x;
if (x>sqn) Q.push(x);
else ++s[x];
--s[i], pre = 0;
}
if (s[i]&1) --s[i], pre = i;
ans += s[i]*i;
if (i*2<=sqn) s[i*2]+=s[i]/2;
else {
REP(j,1,s[i]/2) Q.push(i*2);
}
}
if (pre) Q.push(pre);
for (auto i:q) if (cnt[i]>sqn) Q.push(cnt[i]);
while (Q.size()>1) {
int x = Q.top(); Q.pop();
x += Q.top(); Q.pop();
ans += x, Q.push(x);
}
return ans;
} int main() {
scanf("%d", &n), sqn = sqrt(n);
REP(i,1,n) scanf("%d",a+i),++cnt[a[i]],blo[i]=i/sqn;
REP(i,1,N-1) if (cnt[i]>sqn) q.pb(i);
memset(cnt,0,sizeof cnt);
scanf("%d", &m);
REP(i,1,m) scanf("%d%d",&e[i].l,&e[i].r),e[i].id=i;
sort(e+1,e+1+m);
int ql=1,qr=0;
REP(i,1,m) {
while (ql<e[i].l) upd(a[ql++],-1);
while (qr>e[i].r) upd(a[qr--],-1);
while (ql>e[i].l) upd(a[--ql],1);
while (qr<e[i].r) upd(a[++qr],1);
ans[e[i].id]=calc();
}
REP(i,1,m) printf("%lld\n", ans[i]);
}
E. Cool Slogans
后缀自动机还没学, 以后补了