NOIP2018 模擬賽 10.22
阿新 • • 發佈:2018-12-17
T1 cards
分析
這道題很明顯是一道最長不下降子序列的題,用sort排序+樸素dp可以拿60分,用樹狀陣列優化就能拿100了。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<queue> using namespace std; typedef long long ll; const int MAXN = 1e6+5; int n, m, Ans = -1; int d[MAXN], Dgree[MAXN], f[MAXN], c[MAXN]; int cnt, head[MAXN], To[MAXN*2], Next[MAXN*2]; struct Node { int Atk, Def, Agt; }a[MAXN]; void addEdge(int from,int to) { Dgree[to]++; To[++cnt]=to; Next[cnt]=head[from]; head[from]=cnt; } void toposort() { queue<int> q; for(int i = 1; i <= m; i++) if(!Dgree[i]) q.push(i), d[i]=1; while(q.size()) { int x = q.front(); q.pop(); for(int i = head[x]; i; i = Next[i]) { int y = To[i]; d[y] = d[x] + 1; Dgree[y]--; if(!Dgree[y]) q.push(y); } } } int lowbit(int x) { return x&(-x); } void add(int x,int d) { while(x<=MAXN) { if(d>c[x]) c[x]=d; x+=lowbit(x); } } int ask(int x) { int ret=0; while(x) { if(c[x]>ret) ret=c[x]; x-=lowbit(x); } return ret; } bool comp(const Node &a, const Node &b) { if(a.Agt != b.Agt) return a.Agt <= b.Agt; else if(a.Atk != b.Atk) return a.Atk <= b.Atk; else return a.Def <= b.Def; } int main() { scanf("%d", &m); // cin>>m; int flag=0, ff=0; for(int i = 1; i <= m; i++) { scanf("%d%d%d", &a[i].Atk, &a[i].Def, &a[i].Agt); if(a[i].Def!=0) flag=1; if(a[i].Agt!=0) ff=1; } sort(a+1, a+m+1, comp); if(!flag) { printf("%d", m); //cout<<m<<'\n'; return 0; } if(ff) { for(int i = 1; i <= m; i++) for(int j = 1; j <= m; j++) if(i != j) if(a[i].Atk>=a[j].Atk && a[i].Def>=a[j].Def && a[i].Agt>=a[j].Agt) addEdge(i,j); toposort(); for(int i = 1; i <= m; i++) Ans = max(Ans, d[i]); printf("%d\n", Ans); return 0; } for(int i = 1; i<= m; i++) { int tmp = ask(a[i].Def) + 1; if(Ans < tmp) Ans = tmp; add(a[i].Def, tmp); } printf("%d\n", Ans); retur n 0; }
T2 Pets
分析
這道題我們用拓撲排序來判斷二隊中一個點可以加到一隊中,用f[i][j]表示一隊到i,二隊中到j 是的答案。
#include<cstdio> #include<iostream> #include<cstring> #include<vector> #include<algorithm> #include<cstdlib> #include<queue> #include<bitset> #include<set> #include<deque> #include<climits> using namespace std; typedef long long ll; const int MAXN=1005; int n, m, type[MAXN], du[MAXN], f[MAXN][MAXN]; int s1[MAXN], s2[MAXN], l[MAXN][MAXN], r[MAXN][MAXN]; int map[MAXN][MAXN]; bool topsort(int s[], int kind, int sum) { queue<int> q; int i,j,t = 0; bool mark[MAXN]; memset(mark,0,sizeof(mark)); for(i = 1; i <= n+m; i++) if(type[i]==kind && (!du[i])) { q.push(i); s[++t] = i; mark[i] = 1; } while(q.size()) { int x = q.front(); q.pop(); for(i = 1; i <= n+m; i++) if(!mark[i] && type[i]==kind) { if(map[x][i] && du[i]) du[i]--; if(!du[i]) { q.push(i); mark[i] = 1; s[++t] = i; } } } return t==sum; } int main() { int i,j,x; scanf("%d%d", &n, &m); for(i = 1; i <= n; i++) { for(j = 1; j <= n; j++) scanf("%d", &map[i][j]); type[i] = 2; } for(i=1;i<=m;i++) { scanf("%d", &x); type[x] = 1; } for(i = 1; i <= n; i++) for(j = 1; j <= n; j++) if(type[i]==type[j] && map[i][j]) du[j]++; n -= m; swap(n, m); if(!topsort(s1,1,n) || (!topsort(s2,2,m))) { puts("NO"); return 0; } for(j=1;j<=m;j++) { l[0][j] = r[n+1][j] = 1; for(i = 1; i <= n; i++) l[i][j] = map[s1[i]][s2[j]] && l[i-1][j]; for(i = n; i; i--) r[i][j] = map[s2[j]][s1[i]] && r[i+1][j]; } for(i = 0; i <= n; i++) for(j = 0; j <= m; j++) { if(j) f[i][j] = max(f[i][j], f[i][j-1]); if(i) f[i][j] = max(f[i][j], f[i-1][j]); if(j && l[i][j] && r[i+1][j]) f[i][j] = max(f[i][j], f[i][j-1]+1); } cout << "YES" << " " << f[n][m]; return 0; }