noip题目复习

dp：
背包+ http://www.gdfzoj.com/oj/contest/230/problems/2
奇怪的线段覆盖 http://www.gdfzoj.com/oj/contest/225/problems/2
优化（ln函数的结论）http://www.gdfzoj.com/oj/contest/217/problems/3
http://www.gdfzoj.com/oj/contest/215/problems/2

二分：
http://www.gdfzoj.com/oj/contest/221/problems/2
http://www.gdfzoj.com/oj/contest/211/problems/1
http://www.gdfzoj.com/oj/contest/209/problems/1

搜索：
http://www.gdfzoj.com/oj/contest/212/problems/2
http://www.gdfzoj.com/oj/contest/198/problems/1

图的搜索树：http://www.gdfzoj.com/oj/contest/190/problems/2

倒序求解：
http://www.gdfzoj.com/oj/contest/229/problems/2
http://www.gdfzoj.com/oj/contest/184/problems/3
https://nanti.jisuanke.com/t/18473

结论题：
曼哈顿距离 http://www.gdfzoj.com/oj/contest/217/problems/2
曼哈顿距离 x 线段树 http://www.gdfzoj.com/oj/contest/193/problems/1
mod http://www.gdfzoj.com/oj/contest/211/problems/2
mod x 线段树 http://www.gdfzoj.com/oj/contest/205/problems/1
字符串 http://www.gdfzoj.com/oj/contest/196/problems/3
矩阵 x 期望 http://www.gdfzoj.com/oj/contest/181/problems/3

贪心
http://www.gdfzoj.com/oj/contest/191/problems/1
http://www.gdfzoj.com/oj/contest/184/problems/2

单调栈：http://www.gdfzoj.com/oj/contest/203/problems/2

manacher：http://www.gdfzoj.com/oj/contest/200/problems/1

模板

–>线性筛

for (i=2;i<=n;i++) {        if (!b[i]) prime[tot++] = i, phi[i] = i-1;        for (j=0;j<tot && i*prime[j] <= n;j++) {            b[prime[j] * i] = true;            if (i % prime[j] == 0) {                phi[prime[j] * i] = phi[i] * prime[j];                break            } phi[prime[j] * i] = phi[i] *(prime[j]-1);        }    }

–>线段树

#include <cstdio>#define mid ((l+r)>>1)typedef long long ll;int n,m,op,ql,qr;ll qd;struct node {ll sum,add;} t[100100 << 2];inline int read() {    int ret = 0;    char c = getchar();    while (c > '9' || c < '0') c = getchar();    while (c <= '9' && c >= '0') ret = ret * 10 + c - '0', c = getchar();    return ret;}void down(int l,int r,int k) {    t[k<<1].add += t[k].add;    t[k<<1|1].add += t[k].add;    t[k<<1].sum += t[k].add * (mid - l + 1);    t[k<<1|1].sum += t[k].add * (r - mid);    t[k].add = 0;}void build(int l,int r,int k) {    if (l == r) {t[k].sum = read();  return;};    build(l,mid,k<<1);  build(mid+1,r,k<<1|1);    t[k].sum = t[k<<1].sum + t[k<<1|1].sum;}void add(int l,int r,int k,int L,int R,ll x) {    if (l == L && r == R) {t[k].sum += x * (r-l+1);  t[k].add += x;  return;}  down(l,r,k);    if (R <= mid) add(l,mid,k<<1,L,R,x);  else if (L > mid) add(mid+1,r,k<<1|1,L,R,x);    else add(l,mid,k<<1,L,mid,x),  add(mid+1,r,k<<1|1,mid+1,R,x);    t[k].sum = t[k<<1].sum + t[k<<1|1].sum;}ll query(int l,int r,int k,int L,int R) {    if (l == L && r == R) {return t[k].sum;}  down(l,r,k);    if (R <= mid) return query(l,mid,k<<1,L,R);  else if (L > mid) return query(mid+1,r,k<<1|1,L,R);    else return query(l,mid,k<<1,L,mid) + query(mid+1,r,k<<1|1,mid+1,R);}int main() {    n = read(), m = read();    build(1,n,1);    for (;m--;) {        op = read(), ql = read(), qr = read();        if (op == 2) printf("%lld\n",query(1,n,1,ql,qr));        else add(1,n,1,ql,qr,qd = read());    }}

–>二分图

#include <cstdio>#include <cstring>#include <vector>int n,m,i,q[1010],rt,t1,t2;bool vis[1010];std::vector <int> u[1010];bool f(int v) {    for (int i=0;i<u[v].size();i++)    if (!vis[rt = u[v][i]]) {        vis[rt] = true;        if ((!q[rt]) || f(q[rt])) return q[rt] = v;    } return false;}int main() {    scanf("%d%d",&n,&m);    for (i=1;i<=m;i++) {        scanf("%d%d",&t1,&t2);        u[i].push_back(t1);  u[i].push_back(t2);    }    for (i=1;i<=m;i++) if (memset(vis,0,sizeof(vis)) && !f(i)) break;    printf("%d\n",i - 1);}

–>spfa

#include <cstdio>#include <queue>int N,M,S,i,dis[10010];struct edge {int to,val;  edge *nxt;} e[500500], *head[10010];bool vis[10010];std::queue<int> Q;void spfa(int s) {    dis[s] = 0;    for (Q.push(s);!Q.empty();Q.pop()) {        int u = Q.front();        if (vis[u]) continue;  vis[u] = false;        for (edge *p=head[u];p;p=p->nxt) if (dis[p->to] > dis[u] + p->val) {            dis[p->to] = dis[u] + p->val;            if (!vis[p->to]) Q.push(p->to);        }    }}int main() {    scanf("%d%d%d",&N,&M,&S);    for (i=1;i<=N;i++) dis[i] = 0x7fffffff;    for (i=0;i<M;i++) {        int u,v,w;  scanf("%d%d%d",&u,&v,&w);        head[u] = &(e[i] = (edge) {v,w,head[u]});    } spfa(S);    for (i=1;i<=N;i++) printf("%d ",dis[i]);    putchar('\n');}

–>lca

inline int lca(int a,int b) {    if (dep[a] < dep[b]) swap(a,b);    for (int i=19;~i;i--) if (dep[f[a][i]] >= dep[b]) a = f[a][i];    if (a != b) {for (int i=19;~i;i--) if (f[a][i] != f[b][i]) a = f[a][i], b = f[b][i];  a = f[a][0];}    return a;}

主席树 （考得到算我输）

#include <cstdio>#include <queue>#define N 200100#define mid ((l+r)>>1)using namespace std;int T,n,m,i,rt[N],tot,ans[N],a[N],rnk[N];struct point {int num,ls,rs;} tr[N << 4];priority_queue<pair<int,int> > q;inline int _new() {tr[++tot].num = 0;  return tot;}void add(int now,int old,int k,int l,int r) {    if (l == r) return void(++tr[now].num);    if (k <= mid) add(tr[now].ls = _new(),tr[old].ls,k,l,mid), tr[now].rs = tr[old].rs;    if (k > mid) add(tr[now].rs = _new(),tr[old].rs,k,mid+1,r), tr[now].ls = tr[old].ls;    tr[now].num = tr[tr[now].ls].num + tr[tr[now].rs].num;}int ask(int nowl,int nowr,int k,int l,int r) {    if (l == r) return l;    if (tr[tr[nowr].ls].num - tr[tr[nowl].ls].num >= k) return ask(tr[nowl].ls,tr[nowr].ls,k,l,mid);    return ask(tr[nowl].rs,tr[nowr].rs,k - (tr[tr[nowr].ls].num - tr[tr[nowl].ls].num),mid+1,r);}int main() {    for (scanf("%d",&T);T--;tot = 0) {        scanf("%d%d",&n,&m);        for (i=1;i<=n;i++) {            scanf("%d",a+i);            q.push(make_pair(-a[i],i));        }        for (i=1;i<=n;i++) {            pair<int,int> tmp = q.top();  q.pop();            ans[rnk[tmp.second] = i] = -tmp.first;        }        for (i=1;i<=n;i++) add(rt[i] = _new(),rt[i-1],rnk[i],1,n);        for (int l,r,k;m--;) {            scanf("%d%d%d",&l,&r,&k);            printf("%d\n",ans[ask(rt[l-1],rt[r],k,1,n)]);        }    }}