SCP（大雾）复习板子？

唔

复习一下各种板子？

树状数组

其实忘得差不多了，只会lowbit了qaq

树状数组一的话就正常搞

#include<cstdio>
#define sev en
using namespace std;

#define N 500010

int n,m;
int tree[N];

int lowbit(int x) {
    return x & (-x);
}

void add(int x,int k) {
    while(x <= n) {
        tree[x] += k;
        x += lowbit(x);
    }
}

int query(int x) {
    int ans = 0;
    while(x) {
        ans += tree[x];
        x -= lowbit(x);
    }
    return ans;
}

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++) {
        int x;
        scanf("%d",&x);
        add(i,x);
    }
    for(int i = 1; i <= m; i++) {
        int op,x,k;
        scanf("%d%d%d",&op,&x,&k);
        if(op == 1)
            add(x,k);
        else
            printf("%d\n",query(k) - query(x - 1));
    }
    return 0;
}

BIT1

树状数组二用到了差分

反正是忘了嘤

#include<cstdio>
#define sev en
using namespace std;

#define N 500010

int n,m;
int tree[N];

int lowbit(int x) {
    return x & (-x);
}

void add(int x,int k) {
    while(x <= n) {
        tree[x] += k;
        x += lowbit(x);
    }
}

int query(int x) {
    int ans = 0;
    while(x) {
        ans += tree[x];
        x -= lowbit(x);
    }
    return ans;
}

int main() {
    scanf("%d%d",&n,&m);
    int st = 0;
    for(int i = 1; i <= n; i++) {
        int x;
        scanf("%d",&x);
        add(i,x - st);
        st = x;
    }
    for(int i = 1; i <= m; i++) {
        int op,x,y,k;
        scanf("%d",&op);
        if(op == 1) {
            scanf("%d%d%d",&x,&y,&k);
            add(x,k);
            add(y + 1,-k);
        } else {
            scanf("%d",&x);
            printf("%d\n",query(x));
        }
    }
    return 0;
}

BIT2

并查集

第一遍居然写跪了

（我码之前还表示这要是不会就告辞了脸疼

无论如何反正码风比以前好看了

#include<cstdio>
#define sev en
using namespace std;

#define N 10010

int n,m;
int fa[N];

int get(int x) {
    if(fa[x] == x)
        return x;
    return fa[x] = get(fa[x]);
}

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++)
        fa[i] = i;
    for(int i = 1; i <= m; i++) {
        int op,x,y;
        scanf("%d%d%d",&op,&x,&y);
        if(op == 1)
            fa[get(x)] = get(y);
        else {
            if(get(x) == get(y))
                printf("Y\n");
            else
                printf("N\n");
        }
    }
    return 0;
}

Union Find

线段树

前一阵子刚码过一遍

但仍然出了一些问题

比如需要记住lazy乘16 询问的时候往下传是+= sum也是w

说实在的

现在心情有点差特别差那种有点想离开

真是无趣啊千篇一律的人生碌碌无为的时光蝇营狗苟的未来浑浑噩噩的打转

真烦

#include<cstdio>
#define sev en
using namespace std;

#define N 100010

int n,m;
long long sum[N << 2],lazy[N << 4];
int l[N << 2],r[N << 2];

void build(int now,int L,int R) {
    lazy[now] = 0;
    l[now] = L,r[now] = R;
    if(L == R) {
        scanf("%lld",&sum[now]);
        return ;
    }
    int mid = (L + R) >> 1;
    build(now << 1,L,mid);
    build(now << 1 | 1,mid + 1,R);
    sum[now] = sum[now << 1] + sum[now << 1 | 1];
}

void modify(int now,int L,int R,int k) {
    if(L == l[now] && R == r[now]) {
        lazy[now] += k;
        return ;
    }
    sum[now] += (R - L + 1) * k;
    int mid = (l[now] + r[now]) >> 1;
    if(mid >= R)
        modify(now << 1,L,R,k);
    else if(L > mid)
        modify(now << 1 | 1,L,R,k);
    else {
        modify(now << 1,L,mid,k);
        modify(now << 1 | 1,mid + 1,R,k);
    }
    return ;
}

long long query(int now,int L,int R) {
    sum[now] += (r[now] - l[now] + 1) * lazy[now];
    lazy[now << 1] += lazy[now],lazy[now << 1 | 1] += lazy[now];
    lazy[now] = 0;
    if(L == l[now] && R == r[now])
        return sum[now];
    int mid = (l[now] + r[now]) >> 1;
    if(mid >= R)
        return query(now << 1,L,R);
    else if(L > mid)
        return query(now << 1 | 1,L,R);
    else
        return query(now << 1,L,mid) + query(now << 1 | 1,mid + 1,R);
}

int main() {
    scanf("%d%d",&n,&m);
    build(1,1,n);
    for(int i = 1; i <= m; i++) {
        int op,x,y,k;
        scanf("%d",&op);
        if(op == 1) {
            scanf("%d%d%d",&x,&y,&k);
            modify(1,x,y,k);
        } else {
            scanf("%d%d",&x,&y);
            printf("%lld\n",query(1,x,y));
        }
    }
    return 0;
}

segment tree 1

最短路-dijkstra

唔，虽然有点问题但码的还算可以

为了优化加的vis别忘了就好w

（没加的时候标准版过不了嘤

心情不要太差哦

你看你的私信里有一个明亮的1呢

#include<cstdio>
#include<queue>
#define sev en
using namespace std;

#define N 500010
#define INF 2147483647

int n,m,s,cnt;
int to[N],nxt[N],w[N],head[N];
int dis[N],vis[N];
priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > >q;

void add(int x,int y,int z){
  to[++cnt] = y;
  w[cnt] = z;
  nxt[cnt] = head[x];
  head[x] = cnt;
}

int main(){
  scanf("%d%d%d",&n,&m,&s);
  for(int i = 1;i <= m;i++){
    int x,y,z;
    scanf("%d%d%d",&x,&y,&z);
    add(x,y,z);
  }
  for(int i = 1;i <= n;i++)
    dis[i] = INF;
  dis[s] = 0;
  q.push(make_pair(0,s));
  while(!q.empty()){
    int x = q.top().second;
    q.pop();
    if(vis[x])
    continue;
    vis[x] = 1;
    for(int i = head[x];i;i = nxt[i]){
      int v = to[i];
      if(dis[x] != INF && dis[v] > dis[x] + w[i]){
    dis[v] = dis[x] + w[i];
    q.push(make_pair(dis[v],v));
      }
    }
  }
  for(int i = 1;i <= n;i++)
    printf("%d ",dis[i]);
  return 0;
}

dijkstra

最小生成树-kruskal

难得的一边写过的代码

有点困有点累好闷啊要死了

#include<cstdio>
#include<algorithm>
#define sev en
using namespace std;

#define M 200010
#define N 5010

int n,m;
struct EDGE{
  int x,y,w;
  bool operator <(const EDGE &a)const{
    return w < a.w;
  }
}e[M << 1];
int fa[N],ans,num;

int get(int x){
  if(fa[x] == x)
    return x;
  return fa[x] = get(fa[x]);
}

void kruskal(){
  for(int i = 1;i <= m;i++){
    int l = get(e[i].x),r = get(e[i].y);
    if(l != r){
      fa[l] = r;
      ans += e[i].w;
      num++;
    }
  }
}

int main(){
  scanf("%d%d",&n,&m);
  for(int i = 1;i <= m;i++)
    scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
  sort(e + 1,e + m + 1);
  for(int i = 1;i <= n;i++)
    fa[i] = i;
  kruskal();
  if(num < n - 1)
    printf("orz");
  else
    printf("%d",ans);
  return 0;
}

kruskal

逆序对

重新写了一下逆序对代码跟之前的有一点点不同

大概码风不太一样？更好了？更符合我现在的审美吧

树状数组

#include<cstdio>
#include<algorithm>
#define sev en
using namespace std;

#define N 500010

int n;
int a[N],b[N];
long long ans,tree[N];

int lowbit(int x){
    return x & (-x);
}

void add(int x){
    while(x <= n){
        tree[x]++;
        x += lowbit(x);
    }
}

long long query(int x){
    long long ret = 0;
    while(x){
        ret += tree[x];
        x -= lowbit(x);
    }
    return ret;
}

int main() {
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) {
        scanf("%d",&a[i]);
        b[i] = a[i];
    }
    sort(a + 1,a + n + 1);
    int num = unique(a + 1,a + n + 1) - a - 1;
    for(int i = 1; i <= n; i++)
        b[i] = lower_bound(a + 1,a + num + 1,b[i]) - a;
    for(int i = n; i >= 1; i--) {
        add(b[i]);
        ans += query(b[i] - 1);
    }
    printf("%lld",ans);
    return 0;
}

count inversion 1

归并排序

#include<cstdio>
#define sev en
using namespace std;

#define N 500010

int n;
int a[N],q[N];
long long ans;

void merge(int l,int r) {
    if(l == r)
        return ;
    int mid = (l + r) >> 1;
    merge(l,mid);
    merge(mid + 1,r);
    int i = l,j = mid + 1,k = l;
    while(i <= mid && j <= r) {
        if(a[i] <= a[j])
            q[k++] = a[i++];
        else {
            q[k++] = a[j++];
            ans += (mid - i + 1);
        }
    }
    while(i <= mid)
        q[k++] = a[i++];
    while(j <= r)
        q[k++] = a[j++];
    for(int i = l; i <= r; i++)
        a[i] = q[i];
}

int main() {
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
        scanf("%d",&a[i]);
    merge(1,n);
    printf("%lld",ans);
    return 0;
}

count inversion 2

素数筛法

先写了个素数个数emmm

纯筛没用啥筛法硬刚就好

#include<cstdio>
#include<cmath>
#define sev en
using namespace std;

const int N = 1e8 + 10;

bool f[N];

int main() {
    int n;
    scanf("%d",&n);
    int ans = n - 1;
    for(int i = 2; i <= sqrt(n); i++) {
        if(f[i])
            continue;
        for(int j = i * 2; j <=n; j += i)
            if(!f[j])
                f[j] = 1,ans--;
    }
    printf("%d",ans);
    return 0;
}

prime 1

然后写了个线性筛素数

欧拉筛

一开始写prime数组写到判断是否素数后面了所以是bool类型 de了5分钟气死

emmmm也不清楚到底懂得彻不彻底

希望能记住吧~

#include<cstdio>
#include<cstring>
#define sev en
using namespace std;

const int N = 1e7 + 10;

int n,m;
bool f[N];
int prime[N];
int num;

void Prime() {
    memset(f,true,sizeof(f));
    f[0] = f[1] = 0;
    for(int i = 2; i <= n; i++) {
        if(f[i]) {
            prime[num++] = i;
        }
        for(int j = 0; j < num && i * prime[j] <= n; j++) {
            f[i * prime[j]] = 0;
            if(!(i % prime[j]))
                break;
        }
    }
    return ;
}

int main() {
    scanf("%d%d",&n,&m);
    Prime();
    while(m--) {
        int x;
        scanf("%d",&x);
        if(f[x])
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

prime 2

posted @ 2019-11-11 19:34 ./seven 阅读(190) 评论(0) 编辑收藏举报

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SCP（大雾）复习板子？

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