2017ACM萧山训练第5场(2016 Pacific Northwest – Division 1)
E.Enclosure
做出大小两个凸包，即所有点的凸包和前 k 个点的凸包
按动态凸包的思路，新加入的点会把小凸包上连续的一些点弹出，这些点是一个连续的区间
相当于切掉凸包的一个角，加入一个三角形
若在大凸包上顺时针枚举一个加入的点，这个区间左右端点也是顺时针转的，类似旋转卡壳
切掉部分的面积顺便维护
由于坐标范围较大，用 double 精度会炸
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<stack> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define pa pair<int,int> #define ll long long #define inf 1000000000 #define L(x) (x == 1?tops : x - 1) #define R(x) (x == tops?1 : x + 1) using namespace std; ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n, m, cnt, top, tops; ll sqr(ll x) { return x*x; } struct P{ ll x,y; P(ll _x=0,ll _y=0){x=_x;y=_y;} friend P operator-(P a,P b){ return P(a.x-b.x,a.y-b.y); } friend ll operator*(P a,P b){ return a.x*b.y-a.y*b.x; } friend bool operator<(P a,P b){ if(fabs(a.y-b.y)<0)return a.x<b.x; return a.y<b.y; } friend ll dis2(P a){ return sqr(a.x)+sqr(a.y); } }p[100005], q[100005], qs[100005]; bool cmp(P a,P b) { if(fabs((a-p[1])*(b-p[1]))<0)return dis2(p[1]-a)<dis2(p[1]-b); return (a-p[1])*(b-p[1])>0; } bool cmp2(P a, P b) { if(fabs((a - q[1]) * (b - q[1])) < 0) return dis2(a - q[1]) < dis2(b - q[1]); return (a - q[1]) * (b - q[1]) > 0; } void graham() { top = 0; for(int i=1;i<=cnt;i++) if(p[i]<p[1])swap(p[i],p[1]); sort(p+2,p+cnt+1,cmp); for(int i=1;i<=cnt;i++) { while(top>1&&(q[top]-q[top-1])*(p[i]-q[top-1])<=0) top--; q[++top]=p[i]; } } ll solve() { ll S = 0; qs[tops + 1] = qs[1]; for(int i = 1; i <= tops; i++) S += qs[i] * qs[i + 1]; ll ans = S; int l = 1, r = 1; ll tmp = 0; while((qs[l] - q[1]) * (qs[L(l)] - q[1]) >= 0) { tmp -= (qs[l] - qs[r]) * (qs[L(l)] - qs[l]); l = L(l); } for(int i = 1; i <= top; i++) { while((qs[r] - q[i]) * (qs[R(r)] - q[i]) <= 0) { tmp += (qs[r] - qs[l]) * (qs[R(r)] - qs[l]); r = R(r); } while((qs[R(l)] - q[i]) * (qs[l] - q[i]) < 0) { tmp += (qs[R(l)] - qs[r]) * (qs[l] - qs[r]); l = R(l); } ans = max(ans, S - tmp + (qs[r] - q[i]) * (qs[l] - q[i])); } return ans; } int main() { n = read(); m = read(); cnt = m; for(int i = 1; i <= n; i++) p[i].x = read(), p[i].y = read(); graham(); for(int i = 1; i <= top; i++) qs[i].x = q[i].x, qs[i].y = q[i].y; tops = top; cnt = n; graham(); ll ans = solve(); printf("%lld", ans / 2); if(ans & 1)puts(".5"); else puts(".0"); return 0; } |
G.Maximum Islands
L 的上下左右直接贪心为 W
然后剩下的就是个经典的骑士共存问题，黑白染色最小割
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<stack> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define pa pair<int,int> #define ll long long #define inf 1000000000 using namespace std; ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int cnt = 1, ans, c1, c2; int n, m, ind, T; int id[45][45]; string a[45]; int xx[4] = {1, -1, 0, 0}, yy[4] = {0, 0, 1, -1}; struct edge{ int to, next, v; }e[10000005]; int last[2005],dis[2005],q[2005],h[2005],cur[2005]; void insert(int u,int v,int w) { e[++cnt]=(edge){v,last[u],w};last[u]=cnt; e[++cnt]=(edge){u,last[v],0};last[v]=cnt; } bool bfs() { int head=0,tail=1; for(int i=0;i<=T;i++)h[i]=-1; q[0]=0;h[0]=0; while(head!=tail) { int x=q[head];head++; for(int i=last[x];i;i=e[i].next) if(h[e[i].to]==-1&&e[i].v) { h[e[i].to]=h[x]+1; q[tail++]=e[i].to; } } return h[T]!=-1; } int dfs(int x,int f) { if(x==T)return f; int w,used=0; for(int i=cur[x];i;i=e[i].next) if(h[x]+1==h[e[i].to]&&e[i].v) { w=dfs(e[i].to,min(e[i].v,f-used)); e[i].v-=w;e[i^1].v+=w; if(e[i].v)cur[x]=i; used+=w;if(used==f)return f; } if(!used)h[x]=-1; return used; } void addw(int i, int j) { for(int k = 0; k < 4; k++) { int x = i + xx[k], y = j + yy[k]; if(x < 1 || y < 1 || x > n || y > m)continue; if(a[x][y] == 'C')a[x][y] = 'W'; } } void dfs2(int x, int y) { if(x < 1 || y < 1 || x > n || y > m || a[x][y] != 'L')return; a[x][y] = 'W'; for(int i = 0; i < 4; i++) dfs2(x + xx[i], y + yy[i]); } int main() { n = read(); m = read(); for(int i = 1; i <= n; i++) { cin >> a[i]; a[i] = ' ' + a[i]; } for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(a[i][j] == 'L') addw(i, j); for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(a[i][j] == 'L') { ans++; dfs2(i, j); } for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) id[i][j] = ++ind; T = ind + 1; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(a[i][j] == 'C') { ans++; int u = id[i][j]; if(((i + j) & 1)) { insert(0, u, 1); for(int k = 0; k < 4; k++) { int x = i + xx[k], y = j + yy[k]; if(x < 1 || y < 1 || x > n || y > m || a[x][y] == 'W')continue; int v = id[x][y]; insert(u, v, inf); } } else insert(u, T, 1); } while(bfs()) { for(int i=0;i<=T;i++)cur[i]=last[i]; ans-=dfs(0,inf); } printf("%d\n", ans); return 0; } |
H.Paint
离散化，f(i)表示考虑前 i 个段，最多能染多少
将区间按照右端点排序后树状数组优化 dp
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<stack> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define pa pair<int,int> #define ll long long #define inf 1000000000 using namespace std; ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } ll n, K; ll t[400005], H[400005]; struct data{ ll l, r; }a[200005]; bool cmp(data a, data b) { return a.r < b.r; } void modify(int x, ll v) { for(int i = x; i <= 2 * K; i += i & -i) t[i] = max(t[i], v); } ll query(int x) { if(x == 0)return 0; ll ans = 0; for(int i = x; i; i -= i & -i) ans = max(ans, t[i]); return ans; } int main() { n = read(); K = read(); for(int i = 1; i <= K; i++) { a[i].l = read(), a[i].r = read(); H[2 * i - 1] = a[i].l; H[2 * i] = a[i].r; } sort(H + 1, H + 2 * K + 1); sort(a + 1, a + K + 1, cmp); for(int i = 1; i <= K; i++) { a[i].l = lower_bound(H + 1, H + 2 * K + 1, a[i].l) - H; a[i].r = lower_bound(H + 1, H + 2 * K + 1, a[i].r) - H; } ll ans = 0; for(int i = 1; i <= K; i++) { ll x = query(a[i].l - 1); x += H[a[i].r] - H[a[i].l] + 1; ans = max(ans, x); modify(a[i].r, x); } cout << n - ans << endl; return 0; } |
J.Shopping
从左端点开始，设剩下的钱为 S，每次二分 + rmq找到之后第一个小等于 S 的物品
由于 S 每次减半，复杂度 \(O(30nlogn)\)
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#include<map> #include<cmath> #include<ctime> #include<queue> #include<stack> #include<cstdio> #include<vector> #include<bitset> #include<cstring> #include<iostream> #include<algorithm> #define pa pair<int,int> #define ll long long #define inf 1000000000 using namespace std; ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n, q; int Log[200005]; ll a[200005], f[18][200005]; ll query(int l, int r) { int t = Log[r - l + 1]; return min(f[t][l], f[t][r - (1 << t) + 1]); } ll solve(ll S, int l, int r) { int k = l; while(k <= r) { S = S % a[k]; k++; int tmp = k; int L = tmp, R = r; k = r + 1; while(L <= R) { int MID = (L + R) >> 1; if(query(tmp, MID) > S)L = MID + 1; else R = MID - 1, k = MID; } } return S; } int main() { Log[0] = -1; for(int i = 1; i <= 200000; i++) Log[i] = Log[i >> 1] + 1; n = read(), q = read(); for(int i = 1; i <= n; i++) a[i] = read(); for(int i = 0; i <= 17; i++) for(int j = 1; j + (1 << i) - 1 <= n; j++) if(i == 0)f[i][j] = a[j]; else f[i][j] = min(f[i - 1][j], f[i - 1][j + (1 << (i - 1))]); for(int i = 1; i <= q; i++) { ll S = read(), l = read(), r = read(); printf("%lld\n", solve(S, l, r)); } return 0; } |
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计数器 9850900
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