ACM-ICPC国际大学生程序设计竞赛北京赛区(2017)网络赛
| 编号 | 名称 | 通过率 | 通过人数 | 提交人数 |
| A√水题（队友写的 | Visiting Peking University | 91% | 1122 | 1228 |
| B— | Reverse Suffix Array | 57% | 68 | 119 |
| C 最大子矩阵和 | Matrix | 51% | 182 | 353 |
| D缩点/二分/数位dp | Agent Communication | 11% | 23 | 209 |
| E凸包 | Territorial Dispute | 57% | 327 | 567 |
| F— | Cake | 15% | 15 | 95 |
| G√找规律 | Bounce | 74% | 456 | 609 |
| H— | Polynomial Product | 51% | 17 | 33 |
| I√线段树 | Minimum | 77% | 861 | 1110 |
| J最短路 | Typist's Problem | 57% | 44 | 77 |
#include <iostream> #include <cmath> #include <cstdio> #include <cctype> #include <cstdlib> #include <cstring> #include <climits> #include <set> #include <map> #include <list> #include <queue> #include <stack> #include <string> #include <vector> #include <numeric> #include <sstream> #include <algorithm> #include <functional> using namespace std; typedef long long ll; #define PI acos(-1.0) #define INF 0x3f3f3f3f #define EPS 1e-8 #define MOD 1e9+7 #define max_ 1005 #define maxn 100 int p[max_]; bool vi[max_]; pair<int,int> c[max_]; int main() { int n,m,q; while(~scanf("%d%d",&n,&m)) { int length=0; for(int i=0;i<n;i++) { scanf("%d",&p[i]); vi[i]=true; } scanf("%d",&q); while(q--) { int c; scanf("%d",&c); vi[c]=false; } for(int i=0;i<n;i++) if(vi[i]) { c[length].first=p[i]; c[length++].second=i; } int minn=INF; int aa,bb; for(int i=0;i+m<=length;i++) { int x=c[i].first; for(int j=1;j<m;j++) { if(c[i+j].first+x<minn) { minn=c[i+j].first+x; aa=c[i].second; bb=c[i+j].second; } } } printf("%d %d\n",aa,bb); } return 0; }
#include <iostream> #include <cmath> #include <cstdio> #include <cctype> #include <cstdlib> #include <cstring> #include <climits> #include <set> #include <map> #include <list> #include <queue> #include <stack> #include <string> #include <vector> #include <numeric> #include <sstream> #include <algorithm> #include <functional> using namespace std; typedef long long ll; #define PI acos(-1.0) #define INF 0x3f3f3f3f #define EPS 1e-8 #define MOD 1e9+7 #define max_ 270000 pair<int,int> tree[max_]; int minn,maxx; void add(int rt,int l,int r,int v,int x) { if(l==r) tree[rt].first=tree[rt].second=x; else { int mid=(l+r)>>1; if(mid>=v) add(rt<<1,l,mid,v,x); else add(rt<<1|1,mid+1,r,v,x); tree[rt].first=min(tree[rt<<1].first,tree[rt<<1|1].first); tree[rt].second=max(tree[rt<<1].second,tree[rt<<1|1].second); } } void query(int rt,int l,int r,int L,int R) { if(L>=l&&R<=r) { minn=min(minn,tree[rt].first); maxx=max(maxx,tree[rt].second); } else { int mid=(L+R)>>1; if(mid>=l) query(rt<<1,l,r,L,mid); if(mid<r) query(rt<<1|1,l,r,mid+1,R); } } int main() { int k; int t; scanf("%d",&t); while(t--) { scanf("%d",&k); int e=pow(2,k); for(int i=1;i<=e;i++) { int tmp; scanf("%d",&tmp); add(1,1,e,i,tmp); } int q; scanf("%d",&q); while(q--) { int op; scanf("%d",&op); if(op==1) { int l,r; scanf("%d%d",&l,&r); l++,r++; minn=INF,maxx=-INF; query(1,l,r,1,e); ll ans; if(minn>=0) { ans=minn; ans*=ans; } else if(maxx>=0) { ans=minn; ans*=maxx; } else { ans=maxx; ans*=maxx; } printf("%lld\n",ans); } else { int x,y; scanf("%d%d",&x,&y); add(1,1,e,x+1,y); } } } return 0; }
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int p[105]; int l[105]; struct Node { int day; int cost; }node[105]; int main() { int n, m; while (scanf("%d%d", &n, &m) != EOF) { memset(l, 0, sizeof(l)); for (int i = 0; i < n; i++) { scanf("%d", &p[i]); } int ln,x; scanf("%d", &ln); for (int i = 0; i < ln; i++) { scanf("%d", &l[i]); } sort(l, l + ln); int now = 0; int now_node = 0; for (int i = 0; i < n; i++) { if (i == l[now]&&now<ln) { now++; continue; } node[now_node].day = i; node[now_node].cost = p[i]; now_node++; } int min = 0x3f3f3f3f; int left = 0, right = 0; for (int i = 0; i <= now_node-m; i++) { for (int j = i + 1; j < i + m; j++) { if (node[i].cost + node[j].cost < min) { min = node[i].cost + node[j].cost; left = node[i].day; right = node[j].day; } } } printf("%d %d\n", left,right); } }
#include <iostream> #include <cmath> #include <cstdio> #include <cctype> #include <cstdlib> #include <cstring> #include <climits> #include <set> #include <map> #include <list> #include <queue> #include <stack> #include <string> #include <vector> #include <numeric> #include <sstream> #include <algorithm> #include <functional> using namespace std; typedef long long ll; #define PI acos(-1.0) #define INF 0x3f3f3f3f #define EPS 1e-8 #define MOD 1e9+7 #define max_ 270000 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a; } int main() { ll xx,yy,cc,dd,n,a,b; while(~scanf("%lld%lld",&xx,&yy)) { xx-=1; yy-=1; if(xx<yy) swap(xx,yy); n = xx*yy / gcd(xx,yy); a = n/yy; b = n/xx; if(b!=1) { cc=(xx+yy-1)/yy; dd=a-cc-cc+1; if(b!=2) { n-=(dd%(b-2))*(b-1); dd/=(b-2); n-=dd*(b-2)*(b-1); } n-=(cc-1)*(b-1)*2; } n++; printf("%lld\n",n); } return 0; }
#include <bits/stdc++.h> #define mem(a,b) memset((a),(b),sizeof(a)) #define MP make_pair #define pb push_back #define fi first #define se second #define sz(x) x.size() using namespace std; typedef long long ll; const int INF=0x3f3f3f3f; const ll LLINF=0x3f3f3f3f3f3f3f3f; const double PI=acos(-1.0); const double eps=1e-8; const int MAX=2e5+10; const ll mod=1e9+7; int sgn(double x) { if(fabs(x)<eps) return 0; else return x>0?1:-1; } struct Point { int id; double x,y; Point(){} Point(double a,double b) { x=a; y=b; } void input() { scanf("%lf%lf",&x,&y); } }; typedef Point Vector; Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);} bool operator <(Point a,Point b){return a.x<b.x||(a.x==b.x&&a.y<b.y);} bool operator ==(Point a,Point b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;} double dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;} double cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;} vector<Point> graham(vector<Point> p) { int n,m,k,i; sort(p.begin(),p.end()); p.erase(unique(p.begin(),p.end()),p.end()); n=p.size(); m=0; vector<Point> res(n+1); for(i=0;i<n;i++) { while(m>1&&cross(res[m-1]-res[m-2],p[i]-res[m-2])<=0) m--; res[m++]=p[i]; } k=m; for(i=n-2;i>=0;i--) { while(m>k&&cross(res[m-1]-res[m-2],p[i]-res[m-2])<=0) m--; res[m++]=p[i]; } if(n>1) m--; res.resize(m); return res; } char ans[111]; int main() { int t,n,i; scanf("%d",&t); while(t--) { scanf("%d",&n); vector<Point> v; Point p[111]; for(i=0;i<n;i++) { p[i].input(); p[i].id=i; v.pb(p[i]); } if(n<=2) { puts("NO"); continue; } vector<Point> res=graham(v); mem(ans,0); if(sz(res)==n) { if(n==3) { puts("NO"); continue; } int flag=0; for(i=0;i<sz(res);i++) { ans[res[i].id]='A'+flag; flag=(flag+1)%2; } } else { for(i=0;i<sz(res);i++) { ans[res[i].id]='A'; } for(i=0;i<n;i++) { if(ans[i]!='A') ans[i]='B'; } } ans[n]='\0'; puts("YES"); puts(ans); } return 0; }
#include<bits/stdc++.h> #define rep(i,j,k) for((i)=(j);(i)<=(k);++i) #define per(i,j,k) for((i)=(j);(i)>=(k);--i) using namespace std; typedef long long ll; inline void cmin(ll &x,ll y){if(y<x)x=y;} inline void cmax(ll &x,ll y){if(y>x)x=y;} const ll N = 1000006; const ll inf = 1LL<<60; bool ok[N]; struct edge{ll v,next,w;}e[N]; ll dep[N],last[N],s[N],d[N],fa[N],id1[N],id2[N],K,L,n,i,l,r,u,v,w,cnt,top,stk[N]; ll inline read(){ char ch=getchar();ll z=0,f=1; while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){z=z*10+ch-'0';ch=getchar();} return z*f; } void add(ll u,ll v,ll w){ e[++cnt]=(edge){v,last[u],w};last[u]=cnt; e[++cnt]=(edge){u,last[v],w};last[v]=cnt; } void dfs(ll x,ll y,ll &mx){ if(dep[x] > dep[mx]) mx = x; fa[x] = y; ll i; for(i=last[x];i;i=e[i].next) if(!ok[e[i].v] && e[i].v!=y){dep[e[i].v] = dep[x] + e[i].w; dfs(e[i].v,x,mx);} } ll getlength(ll root,ll &st,ll &ed,ll &len){ dep[st = root] = 0; dfs(root,0,st); len = dep[st]; dep[ed = st] = 0; dfs(st,0,ed); return dep[ed]; } bool cmp1(ll x,ll y){return s[x]+d[x]<s[y]+d[y];} bool cmp2(ll x,ll y){return s[x]-d[x]<s[y]-d[y];} bool solve(ll K){ ll i,j,k,l,p,q,mx=-inf,l1=-inf,r1=inf,l2=-inf,r2=inf; for(k=l=1;k<=top;++k){ j=id1[k]; while(l<=top && s[j]+d[j]-s[id2[l]]+d[id2[l]]>K){i=id2[l++]; cmax(mx,s[i]+d[i]);} if(l>1){ ll mi2=s[id2[1]]-d[id2[1]]; l1=mx+s[top]+L-K; cmin(r1,mi2+s[j]-d[j]-L+K); l2=s[top]+L-K-mi2; cmin(r2,s[j]-d[j]-mx-L+K); } } if(l1>r1) return 0; k = l = top; p = q = 1; rep(i,2,top)if(s[i]*2>=l1+l2&&s[i]*2<=r1+r2){ while(k>0 && s[i]+s[k]>=l1) --k; while(l>0 && s[i]+s[l]>r1) --l; while(p<=top && s[i]-s[p]>=l2) ++p; while(q<=top && s[i]-s[q]>r2) ++q; if(max(q,k+1)<=min(p-1,min(l,i-1))) return 1; } return 0; } int main() { int t; scanf("%d",&t); while (t--) { scanf("%lld",&n); L=1; cnt = 0; rep(i,0,n) last[i] = ok[i] = 0; rep(i,1,n-1){u=read();v=read();add(u,v,1);} ll st,ed,len,l=-1,r=getlength(1,st,ed,len); stk[top = ok[ed] = 1] = ed; for(i=ed;i!=st;i=fa[i]) ok[stk[++top] = fa[i]] = 1; reverse(stk+1,stk+top+1); rep(i,1,top) s[i] = dep[stk[i]]; rep(i,1,top) cmax(l,getlength(stk[i],st,ed,d[i])-1); rep(i,1,top) id1[i] = id2[i] = i; sort(id1+1,id1+top+1,cmp1); sort(id2+1,id2+top+1,cmp2); while(l+1<r){ K = l+r>>1; if(solve(K)) r=K; else l=K; } printf("%lld\n",r); } return 0; }
import java.io.OutputStream; import java.io.IOException; import java.io.InputStream; import java.io.PrintWriter; import java.util.StringTokenizer; import java.math.BigInteger; import java.io.IOException; import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.ArrayList; import java.io.InputStream; /** * Built using CHelper plug-in * Actual solution is at the top */ public class Main { public static void main(String[] args) { InputStream inputStream = System.in; OutputStream outputStream = System.out; InputReader in = new InputReader(inputStream); PrintWriter out = new PrintWriter(outputStream); Task2 solver = new Task2(); int testCount = Integer.parseInt(in.next()); for (int i = 1; i <= testCount; i++) solver.solve(i, in, out); out.close(); } static class Task2 { BigInteger C(int n, int m) { BigInteger ans = BigInteger.valueOf(1); for (int i = 1; i <= m; i++) { ans = ans.multiply(BigInteger.valueOf(n - i + 1)); } for (int i = 2; i <= m; i++) { ans = ans.divide(BigInteger.valueOf(i)); } return ans; } public void solve(int testNumber, InputReader in, PrintWriter out) { int n = in.nextInt(); int[] a = new int[n + 1]; int[] b = new int[n + 1]; for (int i = 1; i <= n; i++) { a[i] = in.nextInt(); b[a[i]] = i; } ArrayList<Integer> arr = new ArrayList<>(); int now = 1; for (int i = 2; i <= n; i++) { if (a[i - 1] != n && (a[i] == n || b[a[i - 1] + 1] > b[a[i] + 1])) { arr.add(now); now = 1; } else { now++; } } arr.add(now); if (arr.size() > 26) { out.println(0); } else { int sz = arr.size(); BigInteger[][] dp = new BigInteger[sz + 1][27]; for (int i = 0; i <= sz; i++) { for (int j = 0; j <= 26; j++) { dp[i][j] = BigInteger.valueOf(0); } } dp[0][0] = BigInteger.valueOf(1); for (int i = 0; i < sz; i++) { for (int j = 0; j < 26; j++) { for (int k = 1; j + k <= 26; k++) { dp[i + 1][j + k] = dp[i + 1][j + k].add(dp[i][j].multiply(C(arr.get(i) + k - 2, k - 1))); } } } // out.println(dp[sz][26]); BigInteger ans = BigInteger.valueOf(0); for (int i = 1; i <= 26; i++) { ans = ans.add(dp[sz][i]); } out.println(ans); } } } static class InputReader { public BufferedReader reader; public StringTokenizer tokenizer; public InputReader(InputStream stream) { reader = new BufferedReader(new InputStreamReader(stream), 32768); tokenizer = null; } public String next() { while (tokenizer == null || !tokenizer.hasMoreTokens()) { try { tokenizer = new StringTokenizer(reader.readLine()); } catch (IOException e) { throw new RuntimeException(e); } } return tokenizer.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } } }
B：java，26^3 dp，http://blog.csdn.net/skywalkert/article/details/51731556（在 cdoj 上也有一个
C：枚举上下界变成最大子段和，变成 dp[i][0..1]，C 直接卡上下界降维。开个数组记录下压缩这一列的最小值
我是枚举上下边界然后单调栈 //
dp[前i列][换了j次][有没有开始选区间]
二维rmq（×）
你用单调栈就不用rmq了
。
区间最小值就是
v[u][d][i]=min(v[u][d-1][i],a[d][i])
然后u d不用存
v[u][d][i]=min(v[u][d-1][i],a[d][i])
然后u d不用存
就和一般的最大子矩阵和一样做
D：http://uoj.ac/problem/298（UOJ原题）抓一条直径，在上面选两个点连，二分答案之后考察可行性，点对的可行区域是一个菱形交（菱形交是百度之星原题）
（如果原本不能满足条件那么就要走这条新边
E：n<3 直接NO；n=3除了共线都是NO；n>=4 取前四个点 讨论一下就行
G：G 是个 TC 原题改，是数论。把这个图对偶一下，考察有多少个正方形被走过。发现是 (m-1)(n-1)/g^2，每个正方形内部一条对角线被走过，有 g-1 个。
J：J直接建图跑最短路，每条 1 边中间加个虚点拆成两个 0.5 就可以 BFS 了
