B.?Connecting Universities

大意: 給定樹, 給定2*k個(gè)點(diǎn), 求將2*k個(gè)點(diǎn)兩兩匹配, 每個(gè)匹配的貢獻(xiàn)為兩點(diǎn)的距離, 求貢獻(xiàn)最大值

單獨(dú)考慮每條邊$(u,v)$的貢獻(xiàn)即可, 最大貢獻(xiàn)顯然是左右兩側(cè)點(diǎn)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;} //headconst 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: 無向有權(quán)圖有重邊自環(huán), 求刪除兩條邊使得s與t不連通, 且兩條邊的邊權(quán)和最小.

先求出任意一條最短路徑, 邊數(shù)顯然不超過$n$, 暴力枚舉這$n$條邊然后再tarjan即可, 復(fù)雜度O(n(m+n))

算是挺簡單的了, 還是打了好久, 一直卡在怎么判斷刪除一條邊后是否連通, 后來發(fā)現(xiàn)tarjan后從s->t經(jīng)過的橋一定是一條鏈, 所以直接dfs就好了, 最后還要注意邊權(quán)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;} //headconst 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

莫隊(duì)一下, 然后將出現(xiàn)次數(shù)小于等于$\sqrt{n}$的暴力合, 其余的用堆合, 復(fù)雜度$O(m\sqrt{n}logn)$, 看了下最優(yōu)解, 好像可以排序一下省去堆從而優(yōu)化掉一個(gè)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; #endifint 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

后綴自動(dòng)機(jī)還沒學(xué), 以后補(bǔ)了

轉(zhuǎn)載于:https://www.cnblogs.com/uid001/p/10574977.html

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