2018 AICCSA Programming Contest
思路:如果存在大于0的交面积的话, 那么肯定能找到一条水平的直线 和 一条垂直的直线,
使得水平直线的左右两边点的个数相等且为n, 垂直直线的左右两边点的个数相等且为n
也就是说不能有点在这两条线上, 否则交面积为0
然后左上角的点和右下角的点配对, 左下角的点和右上角的点配对
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 2e5 + 5, M = 1e5 + 5; const int MOD = 1e9 + 7; pii a[N]; int fac[M]; bool cmp(pii a, pii b) { return a.se < b.se; } void init() { fac[0] = 1; for (int i = 1; i < M; i++) fac[i] = (1LL * fac[i-1] * i) % MOD; } int main() { int T, n; init(); scanf("%d", &T); while(T--) { scanf("%d", &n); for (int i = 1; i <= 2*n; i++) scanf("%d %d", &a[i].fi, &a[i].se); bool f = false; sort(a+1, a+1+2*n); double x = 0, y = 0; if(a[n].fi != a[n+1].fi) x = (a[n].fi + a[n+1].fi) / 2.0; else f = true; sort(a+1, a+1+2*n, cmp); if(a[n].se != a[n+1].se) y = (a[n].se + a[n+1].se) / 2.0; else f = true; int cnt = 0; for (int i = 1; i <= 2*n; i++) if(a[i].fi > x && a[i].se > y) cnt++; if(f) printf("0 "); else printf("%lld ", (1LL * fac[cnt] * fac[n-cnt]) % MOD); } return 0; }
C Function
思路:打表找规律, 发现ai的系数为C(n+1, i) - 1
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 1e6 + 10; const int MOD = 1e9 + 7; int a[N]; LL fac[N], inv[N]; LL q_pow(LL n, LL k) { LL ans = 1; while(k) { if(k&1) ans = (ans * n) % MOD; n = (n * n) % MOD; k >>= 1; } return ans; } void init() { fac[0] = 1; for (int i = 1; i < N; i++) fac[i] = fac[i-1] * i % MOD; inv[N-1] = q_pow(fac[N-1], MOD-2); for (int i = N-2; i >= 0; i--) inv[i] = inv[i+1] * (i+1) % MOD; } LL C(int n, int m) { return fac[n] * inv[m] % MOD * inv[n-m] % MOD; } int main() { int T; init(); scanf("%d", &T); while(T--) { int n; scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); LL ans = 0; for (int i = 1; i <= n; i++) { (ans = ans + (C(n+1, i) - 1) * a[i] % MOD) %= MOD; } printf("%lld ", (ans + MOD) % MOD); } return 0; }
思路:将a数组放到集合里, 方便查找删除
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 1e5 + 100; int a[N], b[N]; multiset<int> s; vector<int> vc; int main() { int T, n, k; scanf("%d", &T); while(T--) { scanf("%d %d", &n, &k); s.clear(); vc.clear(); for (int i = 1; i <= n; i++) scanf("%d", &a[i]), s.insert(a[i]); for (int i = 1; i <= n; i++) scanf("%d", &b[i]); for (int i = 1; i <= n; i++) { multiset<int>:: iterator it = s.lower_bound(b[i]); if(it == s.end() || *it != b[i]) { vc.pb(b[i]); } else s.erase(it); } if((int)vc.size() == 0) puts("YES"); else if((int)vc.size() == 1) { if(*s.begin() - k <= vc[0] && vc[0] <= *s.begin() + k) puts("YES"); else puts("NO"); } else puts("NO"); } return 0; }
F Mirror
思路:对于每个点, 它连向父亲的边只有当它的子树中不能自销的多余部分才会用到
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 1e4 + 5; vector<pii> g[N]; int send[N], rev[N]; LL ans = 0; pii dfs(int u, int o, int w) { pii tmp = {send[u], rev[u]}; for (pii p : g[u]) { if(p.fi != o) { pii pp = dfs(p.fi, u, p.se); tmp.fi += pp.fi; tmp.se += pp.se; } } ans += 1LL * abs(tmp.fi - tmp.se) * w; return tmp; } int main() { int T, n, u, v, w, q; scanf("%d", &T); while(T--) { scanf("%d", &n); for (int i = 1; i <= n; i++) g[i].clear(), send[i] = rev[i] = 0; for (int i = 1; i < n; i++) { scanf("%d %d %d", &u, &v, &w); g[u].pb({v, w}); g[v].pb({u, w}); } scanf("%d", &q); while(q--) { scanf("%d %d", &u, &v); send[u]++; rev[v]++; } ans = 0; dfs(1, 1, 0); printf("%lld ", ans); } return 0; }
思路:背包dp变形
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 1e3 + 5; pii a[N]; int dp[N]; int main() { int T, n, L; scanf("%d", &T); while(T--) { scanf("%d %d", &n, &L); for (int i = 1; i <= n; i++) scanf("%d %d", &a[i].fi, &a[i].se); for (int i = 0; i <= L; i++) dp[i] = 0; for (int i = 1; i <= n; i++) { for (int j = L; j >= a[i].se; j--) { if(j-a[i].se + 1 >= a[i].fi) dp[j] = max(dp[j], dp[j-a[i].se] + 1); } } printf("%d ", dp[L]); } return 0; }
I Tours
思路:二分答案, check时对于每辆bus, 如果它上一天是空闲的, 才能填补今天的空缺, 然后今天原本的车就是昨天需要的车, 不够的拿昨天剩余的补
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 1e5 + 5; vector<int> a[N]; int T, m, n, r, t, tmp[N]; bool check(int x) { int pre = 0; for (int i = 1; i <= n; i++) { if(i == 1) { int tot = 0; for (int j = 1; j <= m; j++) { tmp[j] = (a[j][i] + r - 1) / r; tot += tmp[j]; } if(tot > x) return false; pre = x - tot; } else { int tot = 0; for (int j = 1; j <= m; j++) { if(tmp[j] >= (a[j][i] + r - 1) / r) tmp[j] = (a[j][i] + r - 1) / r; else { if(pre >= (a[j][i] + r - 1) / r - tmp[j]) pre -= (a[j][i] + r - 1) / r - tmp[j]; else return false; tmp[j] = (a[j][i] + r - 1) / r; } tot += (a[j][i] + r - 1) / r; } if(tot > x) return false; pre = x - tot; } } return true; } int main() { scanf("%d", &T); while(T--) { scanf("%d %d %d", &m, &n, &r); for (int i = 1; i <= m; i++) { a[i].clear(); a[i].pb(0); for (int j = 1; j <= n; j++) { scanf("%d", &t); a[i].pb(t); } } int l = 1, r = 5e5, mid = l+r >> 1; while(l < r) { if(check(mid)) r = mid; else l = mid+1; mid = l+r >> 1; } printf("%d ", mid); } return 0; }
思路:2-sat
建边:
对于一条mark的边,
如果其中一点在点覆盖中, 那么另外一点肯定不在点覆盖中
如果其中一点不在点覆盖中, 那么另外一点肯定在点覆盖中
对于一条unmark的边,
如果其中一点在点覆盖中, 那么另外一点不确定
如果其中一点不在点覆盖中, 那么另外一点肯定在点覆盖中
代码:
#include<bits/stdc++.h> using namespace std; const int N=500000; int ins[N],dfn[N],low[N],cnt,sta[N]; int top,v,u; int Next[N],head[N],to[N]; int tot; int scc[N]; int scccnt; void make_list(int u,int v){ Next[++tot]=head[u],head[u]=tot,to[tot]=v; } void tarjan(int x){ ins[x]=dfn[x]=low[x]=++cnt,sta[top++]=x; for(int p=head[x],v=to[p];p;p=Next[p],v=to[p]) if(!dfn[v])tarjan(v),low[x]=min(low[x],low[v]); else if(ins[v])low[x]=min(low[x],dfn[v]); if(low[x]==dfn[x]){ scc[x]=++scccnt,ins[x]=0; while((u=sta[--top])!=x)ins[u]=0,scc[u]=scccnt; } } int main(){ int T; int n,m; int u,v,w; scanf("%d",&T); for(int t=1;t<=T;t++){ scanf("%d%d",&n,&m); memset(head,0,sizeof(int)*(n*2+5)); memset(dfn,0,sizeof(int)*(n*2+5)); top=0; tot=0; memset(scc,0,sizeof(int)*(n*2+5)); memset(low,0,sizeof(int)*(n*2+5)); scccnt=0; memset(ins,0,sizeof(int)*(n*2+5)); cnt=0; memset(sta,0,sizeof(int)*(n*2+5)); for(int i=1;i<=m;i++){ scanf("%d%d%d",&u,&v,&w); if(w){ make_list(u,v+n); make_list(v,u+n); make_list(u+n,v); make_list(v+n,u); } else{ make_list(u+n,v); make_list(v+n,u); } } for(int i=1;i<=2*n;i++){ if(!dfn[i])tarjan(i); } bool ok=1; for(int i=1;i<=n;i++){ ok&=(scc[i]!=scc[i+n]); } if(ok)puts("YES"); else puts("NO"); } return 0; }