2016长乐夏令营 Day7

T1：

找个规律排序一发+树状数组一发O(nlog^2n)

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<stack>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;const int maxn = 2E5 + 20;typedef long long LL;struct Day{int num,t;Day(int _num = 0,int _t = 0) {num = _num; t = _t;}bool operator < (const Day &b) const {return t > b.t;}}day[maxn]; struct Work{int d,r,pos;Work(int _d = 0,int _r = 0,int _pos = 0){d = _d; r = _r; pos = _pos;}bool operator < (const Work &b) const {return d > b.d;}}job[maxn];int n,m,ans[maxn];LL c1[maxn],c2[maxn];void Insert(int pos){for (int j = day[pos].num; j <= m; j += j&-j)c1[j] += 1LL*day[pos].t,c2[j] += 1LL;}bool Judge(int now,int x){LL sumt,sumd; sumt = sumd = 0;for (int j = now; j > 0; j -= j&-j)sumt += c1[j],sumd += c2[j];return sumt - 1LL*job[x].d*sumd >= 1LL*job[x].r;}int getint(){int ret = 0;char ch = getchar();while (ch < '0' || '9' < ch) ch = getchar();while ('0' <= ch && ch <= '9') ret = ret*10 + ch - '0',ch = getchar();return ret;}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);freopen("test.txt","w",stdout);    #elsefreopen("work.in","r",stdin);freopen("work.out","w",stdout);#endif//scanf("%d%d",&n,&m);n = getint(); m = getint(); for (int i = 1; i <= m; i++) {int x = getint(); //scanf("%d",&x);day[i] = Day(i,x);}sort(day + 1,day + m + 1);for (int i = 1; i <= n; i++) {int x,y; x = getint(); y = getint();//scanf("%d%d",&x,&y);job[i] = Work(x,y,i); }sort(job + 1,job + n + 1);int tail = 1;for (int i = 1; i <= n; i++) {while (tail <= m && day[tail].t > job[i].d) Insert(tail++);int L,R; L = 1; R = m+1;while (R - L > 1) {int mid = (L+R) >> 1;if (Judge(mid,i)) R = mid;else L = mid;}int Ans;if (Judge(L,i)) Ans = L;else Ans = R;if (Ans == m + 1) Ans = 0;ans[job[i].pos] = Ans;}for (int i = 1; i <= n; i++) printf("%d ",ans[i]);return 0;}

T2：

画个图易发现两条路径若相交，必与LCA有关，两两枚举冲突路线，建边，做一发最大独立子集

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<stack>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;const int maxn = 1E5 + 10;const int S = 0;const int T = 4901;const int INF = ~0U>>1;typedef long long LL;struct E{int from,to,cap,flow;E(int _from = 0,int _to = 0,int _cap = 0,int _flow = 0) {from = _from; to = _to; cap = _cap; flow = _flow;}}edgs[maxn*10];int n,L,P,cnt,tot,fa[maxn][20],Deep[maxn],cur[maxn];LL ans = 0;vector <int> v[maxn];vector <int> v2[maxn];queue <int> Q;void dfs(int x,int from){for (int i = 1; i < 20; i++)fa[x][i] = fa[fa[x][i-1]][i-1];for (int i = 0; i < v[x].size(); i++) {int to = v[x][i];if (to == from) continue;Deep[to] = Deep[x] + 1;fa[to][0] = x;dfs(to,x);}}int LCA(int p,int q){if (Deep[p] < Deep[q]) swap(p,q);int Log; for (Log = 0; Deep[p] - (1<<Log) > 0; Log++); --Log;for (int j = Log; j >= 0; j--)if (Deep[p] - (1<<j) >= Deep[q])p = fa[p][j];if (p == q) return p;for (int j = Log; j >= 0; j--)if (fa[p][j] != fa[q][j])p = fa[p][j],q = fa[q][j];return fa[p][0]; }struct B{int a,b,w,num,lca;B(int _a = 0,int _b = 0,int _w = 0,int _num = 0) {a = _a; b = _b; w = _w; num = _num; lca = LCA(_a,_b);}}lpf[maxn],pyz[maxn];bool arg(int x,int y){B p = lpf[x],q = pyz[y];if (Deep[p.lca] > Deep[q.lca]) swap(p,q);/*if (q.a != q.lca) {int Lca = LCA(p.a,q.lca);if (Lca == q.lca && p.a != q.lca) {Lca = LCA(p.a,q.a);if (Lca == p.a || Lca == q.a) return 1;}Lca = LCA(p.b,q.lca);if (Lca == q.lca && p.b != q.lca) {Lca = LCA(p.b,q.a);if (Lca == p.b || Lca == q.a) return 1;}}if (q.b != q.lca) {int Lca = LCA(p.a,q.lca);if (Lca == q.lca && p.a != q.lca) {Lca = LCA(p.a,q.b);if (Lca == p.a || Lca == q.b) return 1;}Lca = LCA(p.b,q.lca);if (Lca == q.lca && p.b != q.lca) {Lca = LCA(p.b,q.b);if (Lca == p.b || Lca == q.b) return 1;}}*/int Lca = LCA(q.lca,p.a);if (Lca == q.lca) return 1;Lca = LCA(q.lca,p.b);if (Lca == q.lca) return 1;return 0;}void Add(int from,int to,int cap){v2[from].push_back(cnt); edgs[cnt++] = E(from,to,cap,0);v2[to].push_back(cnt); edgs[cnt++] = E(to,from,0,0);}bool BFS(){for (int i = 1; i <= tot; i++) Deep[i] = 0;Deep[T] = 0; Deep[S] = 1; Q.push(S);while (!Q.empty()) {int k = Q.front(); Q.pop();for (int i = 0; i < v2[k].size(); i++) {E e = edgs[v2[k][i]];if (e.cap == e.flow) continue;if (Deep[e.to]) continue;Deep[e.to] = Deep[k] + 1;Q.push(e.to);}}return Deep[T];}int Dicnic(int x,int a){if (x == T) return a;int flow = 0;for (int &i = cur[x]; i < v2[x].size(); i++) {E &e = edgs[v2[x][i]];if (e.cap == e.flow) continue;if (Deep[e.to] != Deep[x] + 1) continue;int f = Dicnic(e.to,min(a,e.cap - e.flow));if (f > 0) {flow += f;e.flow += f;edgs[v2[x][i]^1].flow -= f;a -= f;if (!a) return flow;}}if (!flow) Deep[x] = -1;return flow;}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);    #elsefreopen("bus.in","r",stdin);freopen("bus.out","w",stdout);#endifcin >> n >> L >> P; for (int i = 1; i < n; i++) {int x,y; scanf("%d%d",&x,&y);v[x].push_back(y); v[y].push_back(x);}Deep[1] = 1; dfs(1,0);for (int i = 1; i <= L; i++) {int x,y,w; scanf("%d%d%d",&x,&y,&w);lpf[i] = B(x,y,w,++tot);ans += 1LL*w;}for (int i = 1; i <= P; i++) {int x,y,w; scanf("%d%d%d",&x,&y,&w);pyz[i] = B(x,y,w,++tot);ans += 1LL*w;}for (int i = 1; i <= L; i++)for (int j = 1; j <= P; j++)if (arg(i,j))Add(lpf[i].num,pyz[j].num,INF);for (int i = 1; i <= L; i++) Add(S,lpf[i].num,lpf[i].w);for (int i = 1; i <= P; i++) Add(pyz[i].num,T,pyz[i].w);LL flow = 0;while (BFS()) {for (int i = 1; i <= tot; i++) cur[i] = 0;cur[S] = cur[T] = 0;flow += 1LL*Dicnic(S,INF); }cout << ans - flow;return 0;}

T3：

树dp  f[i][j] : i的子树中，是(否)已经完全填充，若是(否)，还可以向上(下)扩展j个单位

显然，0 <= j <= 100，这个答案的优劣程度具有显然的单调性，即，如果f[i][j+1] < f[i][j] ，那么f[i][j]可以等于f[i][j+1]  未完全填满的情况也一样

我们让-100 <= j <= 100 (开一个第二维201的数组，两种0是要分开讨论的)

如果选i  f[i][w[i]] = 1 + ∑min(f[son][j])   -w[i] <= j <= 100

如果不选i f[i][j] = 

if (j >= 0)  f[k][j+1] + ∑min(f[son][x])  (son != k,-j <= x <= 100)

if (j <= 0)  f[k][j+1] + ∑min(f[son][x]) (son != k,j < x <= 100)

k通过枚举和最大值、次大值的特判

min可以用上述单调性O(1)维护与查询

对于f中的非法状态，直接赋值INF，f数组开LL，这样方便编写

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<stack>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;const int maxn = 1E5 + 10; const int INF = maxn;typedef long long LL;LL sum[210],f[maxn][210];int n,m,t,w[maxn],Maxw[maxn];vector <int> v[maxn];void Work(int x,int from){memset(sum,0,sizeof(sum));for (int i = 0; i < v[x].size(); i++) {int to = v[x][i];if (to == from) continue;for (int j = 0; j <= 201; j++) sum[j] += f[to][j];}for (int i = 0; i <= 200; i++) {f[x][i] = INF;if (i > 100)for (int j = 0; j < v[x].size(); j++) {int to = v[x][j];if (to == from) continue;//if (f[to][i+1] == INF) continue;f[x][i] = min(f[x][i],f[to][i+1] + sum[202-i] - f[to][202-i]);}else f[x][i] = sum[i+1];}f[x][201] = INF;}void dfs(int x,int from){if (v[x].size() == 1 && from) {for (int i = 101; i <= w[x] + 101; i++) f[x][i] = 1;for (int i = w[x] + 102; i <= 201; i++) f[x][i] = INF;for (int i = 0; i <= 100; i++) f[x][i] = 0;return;}for (int i = 0; i < v[x].size(); i++) {int to = v[x][i];if (to == from) continue;dfs(to,x);}Work(x,from);LL tot = 1LL + sum[101-w[x]];f[x][w[x]+101] = min(f[x][w[x]+101],tot);for (int i = 200; i >= 0; i--) f[x][i] = min(f[x][i],f[x][i+1]);}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);    #elsefreopen("city.in","r",stdin);freopen("city.out","w",stdout);#endifcin >> t;while (t--) {cin >> n;for (int i = 1; i <= n; i++) scanf("%d",&w[i]);for (int i = 1; i < n; i++) {int x,y; scanf("%d%d",&x,&y);v[x].push_back(y); v[y].push_back(x);} dfs(1,0);printf("%d\n",f[1][101]);for (int i = 1; i <= n; i++) v[i].clear();}return 0;}