信息学竞赛算法指北

排序

STL排序

#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
int main(){
    const int N = 5;
    int a [N] = {12,35,99,18,76};
    sort(a,a+5);
    for(int i = 0;i<N;i++){
        cout<<a[i]<<' ';
    }
    return 0;
}

桶排序

#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
int main(){
	int a = 1,b=100,c=5;
	int srt[100001] = {0};
	//cin>>a>>b>>c;
	srt[a]++;
	srt[b]++;
	srt[c]++;
	for(int i = 1;i<=100000;i++){
		while(srt[i]){
			cout<<i<<' ';
			srt[i]--;
		}
	}
	return 0;
}

冒泡排序

#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
int main(){
    const int N = 5;
    int a [N] = {12,35,99,18,76};

    for(int i = 0;i<N;i++){
        for(int j = 0;j<N-i;j++){
            if(a[j]>a[j+1])
                swap(a[j],a[j+1]);
        }
    }
    for(int i = 0;i<N;i++){
        cout<<a[i]<<' ';
    }
    return 0;
}

快速排序

#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
const int N = 5;
int a [N] = {12,35,99,18,76};
void qs(int l,int r){
    if(l>=r)
        return ;
    int t = a[l];
    int i = l,j=r;
    while(i!=j){
        //必须先右后左
        while(a[j]>=t&&i<j)
            j--;
        while(a[i]<=t&&i<j)
            i++;
        if(i<j)
            swap(a[i],a[j]);
    }
    a[l] = a[i];
    a[i] = t;
    qs(l,i-1);
    qs(i+1,r);
    return ;
}
int main(){
    qs(0,N-1);
    for(int i = 0;i<N;i++){
        cout<<a[i]<<' ';
    }
    return 0;
}

数论

逆元

• 自己瞎写的，可能是错的

//这个真的太慢了，请看下面那个
#include<bits/stdc++.h>
using namespace std;
const int A = 3,p=11;
unsigned long long inv = 1;
int main(){
	int i;
	for(i = 0;;i++){
		inv = ((i*p+1)/A)%p;
		if((A*inv)%p==1)
			break;
	}
	cout << i << ' ' << inv;
	return 0;
}

• 现在才是真正的求逆元，用了exgcd

#include<cstdio>
#include<iostream>
using namespace std;
#define pr pair<int ,int >
pr exgcd(int a,int b){
	if(b==0) return pr(1,0);
	pr tmp = exgcd(b,a%b);
	return pr(tmp.second,tmp.first-a/b*tmp.second);
}

int main(){
	int a,b;
	cin>>a>>b;
	int t = exgcd(a,b).first;
	if(t < 0)
		t = t + b;
	cout <<t;
	return 0;
}

欧拉函数

int phi(int n){//对n求欧拉
    int Ans = n;
    for(int i = 2;i<=sqrt(n);i++){//找出每一个因数
        if(n%i == 0){//是因数
            Ans = Ans / i * (i-1);//欧拉函数分解求法
            while(n%i==0)
                n = n / i;//去除该因子
        }
    }
    if(n>1)
        Ans = Ans / n * (n-1);//最后还有因子
    return Ans;
}

图论

最短路

SPFA (单源)

//Code From 算法竞赛进阶指南
#include <bits/stdc++.h>
using namespace std;
const int N = 100010, M = 1000010;
int head[N],ver[M],edge[M],Next[M],d[N];//d是从初始顶点到其他点的距离,ver使出度,edge是路径长度,Next是下一个该顶点的出度，即原来的第二个出度
int n,m,tot;
queue<int> q;
bool v[N];//顶点是否已在队列中

void add(int x,int y,int z){
    //邻接表的添加
    ver[++tot] = y;//ver是终止顶点，tot是总路径数，y是终止顶点
    edge[tot] = z;//edge是路径长度
    Next[tot] = head[x];//Next是下一个该顶点的出度，即原来的第二个出度
    head[x] = tot;//将原来的第二个出度替换为当前添加的出度，原来的第二个出度存放在现在的第二个出度的下一个出度中
}
void SPFA(){
    memset(d,0x3f,sizeof(d));//默认将d最大化，0x3f 这里不解释,反正很大
    memset(v,0,sizeof(v));
    d[1] = 0;//1到1的距离为1
    v[1] = 1;//1号顶点在队列中,与下面一句组合使用使队列中同一顶点只出现一次
    q.push(1);//这里面的1是出发顶点
    while(q.size()){//队列不为空时
        int x = q.front();//取出队首
        q.pop();//删除队首
        v[x] = 0;//代表x已被移除队列
        for(int i = head[x];i != 0;i=Next[i]){//遍历队首所能到达的每一条边
            int y = ver[i];//到达的边
            int z = edge[i];//边长
            if(d[y]>d[x]+z){//判断是否更改"最小距离"
                d[y] = d[x] + z;
                if(!v[y]){
                    q.push(y);//将y顶点加入队列
                    v[y] = 1;//代表y顶点已被移出队列
                }
            }

        }
    }
}
int main(){
    cin>>n>>m;
    for(int i = 1;i<=m;i++){//读入m条边
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        add(x,y,z);
    }
    SPFA();
    for(int i = 1;i<=n;i++){
        cout<<d[i]<<endl;
    }
    return 0;
}

//source:5 5 2 3 2 1 2 -3 1 5 5 4 5 2 3 4 3
//ans:0 -3 -1 2 4

Floyd-Wallshall (整张图)

#include <bits/stdc++.h>
#include <conio.h>
using namespace std;
int e[200][200];
int n,m;
void Floyd_Wallshall(){
    for(int k = 1;k<=n;k++){//三层循环通过k来求得当前状态下i->j的路径最小值
        for(int i = 1;i<=n;i++){
            for(int j = 1;j<=n;j++){
                if(e[i][j] > e[i][k] + e[k][j])
                    e[i][j] = e[i][k] + e[k][j];
            }
        }
    }
    return ;
}
int main(){
    cin>>n>>m;
    for(int i = 1;i<=n;i++){
        memset(e[i],0x3f,sizeof(e[i]));
        e[i][i] = 0;//初始化
    }
    for(int i = 1;i<=m;i++){
        int x,y,z;
        scanf("%d %d %d",&x,&y,&z);//直接读入
        e[x][y] = z;//直接读入
    }
    Floyd_Wallshall();
    for(int i = 1;i<=n;i++){//输出
        for(int j = 1;j<=n;j++){
            printf("%d ",e[i][j]);
        }
        printf("
");
    }
    return 0;
}

负环

• SPFA

//Luogu P3385 【模板】负环
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <ctime>

using namespace std;


const int MaxN = 2003,MaxM= 3005;
int cnt[MaxN];
int tot,T,N,M,d[MaxN];
bool v[MaxN];
int first[MaxN],ver[MaxM * 2],Next[MaxM * 2],wei[MaxM * 2];


void read(int x,int y,int z){
    Next[++tot] = first[x];
    first[x] = tot;
    ver[tot] = y;
    wei[tot] = z;
}
void spfa(){
    int x,y,z;
    memset(d,0x3f,sizeof(d));
    memset(v,0,sizeof(v));
    memset(cnt,0,sizeof(cnt));
    d[1] = 0;
    v[1] = 1;
    queue <int> q;
    q.push(1);
    cnt[1] = 1;
    while(q.size()){
        x = q.front();
        q.pop();
        v[x] = false;
        for(int i = first[x];i;i=Next[i]){
            y = ver[i];
            z = wei[i];
            if(d[y]>d[x]+z){
                d[y] = d[x] + z;
                cnt[y] = cnt[x]+1;
                if(cnt[y]>N){
                    cout<<"YE5"<<endl;
                    return ;
                }
                if(!v[y])
                    q.push(y),v[y] = 1;
            }
        }
    }
    cout<<"N0"<<endl;
    return ;
}

void calc(){
    memset(first,0,sizeof(first));
    memset(Next,0,sizeof(Next));
    tot = 0;
    scanf("%d %d",&N,&M);
    int t1,t2,t3;
    for(int i = 1;i<=M;i++){
        scanf("%d %d %d",&t1,&t2,&t3);
        if(t3>=0)
            read(t2,t1,t3);
        read(t1,t2,t3);
    }
    spfa();
}

int main()
{
    scanf("%d",&T);
    while(T--)
        calc();
    return 0;
}

割点割边

Tarjan判割点

#include <iostream>
#include <cstdio>

using namespace std;

int n,m;
int first[1000],next[10000],ver[10000];
int tot = 0;
int dfn[1000],low[1000],num=1;
int flag[1000];

void tarjan(int node,int father){
    dfn[node] = low[node] = num++;
    int child = 0;
    for(int i = first[node];i;i=next[i]){
        int togo = ver[i];
        if(dfn[togo] == 0){
            child++;
            tarjan(togo,node);
            low[node] = min(low[node],low[togo]);
            if(node != 1 && low[togo] >= dfn[node]){
                flag[node] = 1;
            }
            else if(node == 1 && child == 2){
                flag[1] = 1;
            }
        }
        else if (togo != father){
            low[node] = min(low[node],low[togo]);
        }
    }
    return ;
}

void read(int x,int y){
    next[tot] = first[x];
    first[x] = tot;
    ver[tot] = y;
    tot++;
    return;
}

int main()
{
    scanf("%d %d",&n,&m);
    int t1,t2;
    for(int i = 1;i<=m;i++){
        scanf("%d %d",&t1,&t2);
        read(t1,t2);
        read(t2,t1);
    }
    tarjan(1,1);
    for(int i = 1;i<=n;i++){
        if(flag[i])
            cout<<i<<' ';
    }
    return 0;
}
//6 7 1 4 1 3 4 2 3 2 2 5 2 6 5 6
//2

并查集

• Luogu P3367

#include <iostream>
#include <cstdio>

using namespace std;

#define maxn 10002
int N,M;
int fa[maxn];

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

int main()
{
    scanf("%d%d",&N,&M);
    for(int i = 1;i<=N;i++){
        fa[i] = i;
    }
    int t1,t2,t3;
    for(int i = 1;i<=M;i++){
        scanf("%d %d %d",&t1,&t2,&t3);
        if(t1 == 1){
            fa[f(t2)] = f(t3);
        }else {
            if(f(t2) == f(t3)){
                printf("Y
");
            }else{
                printf("N
");
            }
        }
    }
    return 0;
}

二分图匹配

• Luogu P3386
• 以下为未优化代码

用时: 2533ms / 内存: 7100KB

#include <iostream>
#include <bitset>
#include <cstdio>

using namespace std;

#define maxn 2002
#define maxm 1002001
#define itn int

bitset <maxn> book;

int tot,N,M,sum;
int match[maxn],first[maxn],ver[maxm],next[maxm];


void read(int x,int y){
    next[++tot] = first[x];
    first[x] = tot;
    ver[tot] = y;
}
void clean(){
    bitset <maxn> nul;
    book = nul;
}
bool dfs(int nd){
    for(int i = first[nd];i;i=next[i]){
        int v = ver[i];
        if(!book[v]){
            book[v]=1;
            if(match[v] == 0 || dfs(match[v])){
                match[v] = nd;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    int tN,t1,t2;
    scanf("%d %d %d",&N,&tN,&M);

    for(int i = 1;i<=M;i++){
        scanf("%d %d",&t1,&t2);
        if(t1>N||t2>tN){
            i--,M--;
            continue;
        }
        read(t1,t2);
    }
    for(int i = 1;i<=N;i++){
        clean();
        if(dfs(i))
            sum++;
    }
    printf("%d",sum);
    return 0;
}

• 以下为优化代码

用时: 2213ms / 内存: 2992KB

#include <iostream>
#include <bitset>
#include <cstdio>

using namespace std;

#define maxn 2002
#define maxm 1002001
#define itn int

bitset <maxn> book;

int tot,N,M,sum;
int match[maxn],first[maxn],ver[maxm],next[maxm];

inline int qr(){
    char ch;
    int ret = 0,f = 1;
    ch = getchar();
    while(!isdigit(ch)){
        if(ch == '-')
            f = -1;
        ch = getchar();
    }	
    while(isdigit(ch)){
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return f * ret;
}


inline void read(int x,int y){
    next[++tot] = first[x];
    first[x] = tot;
    ver[tot] = y;
}
inline void clean(){
    bitset <maxn> nul;
    book = nul;
}
bool dfs(int nd){
    for(register int i = first[nd];i;i=next[i]){
        int v = ver[i];
        if(!book[v]){
            book[v]=1;
            if(match[v] == 0 || dfs(match[v])){
                match[v] = nd;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    int tN,t1,t2;
    //scanf("%d %d %d",&N,&tN,&M);	
    N = qr();
    tN = qr();
    M = qr();
    
    for(register int i = 1;i<=M;i++){
        //scanf("%d %d",&t1,&t2);
        t1 = qr();
        t2 = qr();
        if(t1>N||t2>tN){
            i--,M--;
            continue;
        }
        read(t1,t2);
    }
    for(register int i = 1;i<=N;i++){
        clean();
        if(dfs(i))
            sum++;
    }
    printf("%d",sum);
    return 0;
}

强联通分量

Tarjan

• Luogu P2863

#include <iostream>
#include <cstdio>

using namespace std;

#define maxn 10002
int N,M;
int fa[maxn];

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

int main()
{
    scanf("%d%d",&N,&M);
    for(int i = 1;i<=N;i++){
        fa[i] = i;
    }
    int t1,t2,t3;
    for(int i = 1;i<=M;i++){
        scanf("%d %d %d",&t1,&t2,&t3);
        if(t1 == 1){
            fa[f(t2)] = f(t3);
        }else {
            if(f(t2) == f(t3)){
                printf("Y
");
            }else{
                printf("N
");
            }
        }
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <bitset>
#include <stack>


using namespace std;

#define MaxN 10002
#define MaxM 50002
int N,M,tot,id,cnt,ans;
bitset <MaxN> ins;
bitset <MaxN> use;
int color[MaxN] ;
int dfn[MaxN],num[MaxN];
int first[MaxN],ver[MaxM],Next[MaxM];
stack <int> s;

int qr() {
    char ch = getchar();
    int ans = 0;
    while(!isdigit(ch))
        ch = getchar();
    while(isdigit(ch)) {
        ans = ans * 10 + ch - '0';
        ch = getchar();
    }
    return ans;
}

void read(int x,int y) {
    Next[++tot] = first[x];
    first[x] = tot;
    ver[tot] = y;
}

void tarjan(int nd) {
    ins[nd] = 1;
    s.push(nd);
    dfn[nd] = num[nd] = ++id;
    for(int i = first[nd]; i; i=Next[i]) {
        int access = ver[i];
        if(!dfn[access]) {
            tarjan(access);
            num[nd] = min(num[nd],num[access]);
        } else if(ins[access]) {
            num[nd] = min(num[nd],dfn[access]);
        }
    }
    if(dfn[nd] == num[nd]){
        cnt++;
        int v;
        do{
            color[cnt]++;
            v = s.top();
            s.pop();
            ins[v] = 0;
            use[v] = 1;
        }while(nd != v);
    }
}

int main() {
    N = qr();
    M = qr();
    int t1,t2;
    for(int i = 1; i<=M; i++) {
        t1 = qr();
        t2 = qr();
        read (t1,t2);
    }
    for(int i = 1; i<=N; i++) {
        if(!use[i])
            tarjan(i);
    }
    for(int i = 1;i<=cnt;i++){
        if(color[i]>1)
            ans++;
    }
    cout<<ans;
    return 0;
}

动态规划

0/1背包

for (int i = 1; i <= n; ++i) {
	for (int j = 0; j <= m; ++j) {
		if (j < w[i])
			f[i][j] = f[i - 1][j];
		else
			f[i][j] = max(f[i - 1][j], f[i - 1][j - w[i]] + v[i]);
	}
}

完全背包

#include <iostream>

using namespace std;

int T,M;

struct drug{
    int time;
    int value;
}a[105];
int f[1005];


int main()
{
    ios::sync_with_stdio(false);
    cin>>T>>M;
    int t1,t2;
    for(int i = 1;i<=M;i++){
        cin>>a[i].time>>a[i].value;
    }
    for(int i = 1;i<=M;i++){
        for(int j = a[i].time;j<=T;j++){
            f[j]=max(f[j],f[j-a[i].time]+a[i].value);
        }
    }
    cout<<f[T];
    return 0;
}

数据结构

线段树

线段树讲解

单点修改多点查询

• ACOJ 12222

#include <iostream>
#include <cstdio>
#include <cstdlib>

using namespace std;

#define ll long long
#define l(p) p<<1
#define lt(p) t[p<<1]
#define r(p) p<<1|1
#define rt(p) t[p<<1|1]
#define maxn 100000
#define Mid int mid = (t[p].l+t[p].r)>>1
#define Luogu

#ifdef Luogu
#define read scanf("%d",&M),build(1,1,N);
#endif // Luogu

#ifdef Acoj
#define read build(1,1,N),scanf("%d",&M);
#endif // Acoj

struct node{
    int l,r;
    ll w,f;
}t[maxn<<2+3];

void build(int p,int l,int r){
    t[p].l = l;
    t[p].r = r;
    if(l == r){
        //int t;
        //scanf("%d",&t);
        scanf("%lld",&t[p].w);
        return;
    }
    int mid = (l+r)>>1;
    build(l(p),l,mid);
    build(r(p),mid+1,r);
    t[p].w = lt(p).w+rt(p).w;
}

void spread(int p){
    if(t[p].f){
        lt(p).w += t[p].f*(lt(p).r-lt(p).l+1);
        rt(p).w += t[p].f*(rt(p).r-rt(p).l+1);
        lt(p).f += t[p].f;
        rt(p).f += t[p].f;
        t[p].f = 0;
    }
}

void change(int p,int l,int r,int z){
    if(t[p].l>=l&&t[p].r<=r){
        t[p].w += (ll)z*(t[p].r-t[p].l+1);
        t[p].f += z;
        return ;
    }
    spread(p);
    Mid;
    if(l<=mid) change(l(p),l,r,z);
    if(r> mid) change(r(p),l,r,z);
    t[p].w = lt(p).w+rt(p).w;
}
ll ask(int p,int l,int r){
    if(t[p].l>=l&&t[p].r<=r){
        return t[p].w;
    }
    spread(p);
    ll ans = 0;
    Mid;
    if(l<=mid) ans+=ask(l(p),l,r);
    if(r> mid) ans+=ask(r(p),l,r);
    return ans;
}

int N,M;
int main()
{
    scanf("%d",&N);
    read
    int t1,t2,t3;
    for(int i = 1;i<=M;i++){
        scanf("%d",&t1);
        if(t1 == 1){
            scanf("%d %d %d",&t1,&t2,&t3);
            change(1,t1,t2,t3);
        }else {
            scanf("%d %d",&t2,&t3);
            cout<<ask(1,t2,t3)<<endl;
        }
    }
    return 0;
}

树状数组

单点修改

• LuoguP3374

#include <iostream>
#include <cstdio>

using namespace std;

int A[500002];
int N,M;
int lowbit(int a){
    return a&(-a);
}
void add(int a,int b){
    while(a<=N){
        A[a]+=b;
        a += lowbit(a);
    }
}
int Find(int a){
    int ans = 0;
    while(a>0){
        ans += A[a];
        a -= lowbit(a);
    }
    return ans;
}
int main()
{
    int t1,t2,t3;
    scanf("%d %d",&N,&M);
    for(int i = 1;i<=N;i++){
        scanf("%d",&t1);
        add(i,t1);
    }
    for(int i = 1;i<=M;i++){
        scanf("%d",&t1);
        if(t1 == 1){
            scanf("%d %d",&t2,&t3);
            add(t2,t3);
        } else {
            scanf("%d %d",&t2,&t3);
            cout<<Find(t3)-Find(t2-1)<<endl;
        }
    }
    return 0;
}

ST表

• 这个是真的迷
• Luogu P3865

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

int a[100005][40];

inline int qr(){
    char ch;
    int ret=0;
    ch = getchar();
    while(!isdigit(ch))
        ch = getchar();
    while(isdigit(ch)){
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret;
}
int main()
{
    int N,M;
    N=qr();M=qr();
    for(int i = 1;i<=N;i++){
        a[i][0] =  qr();
    }
    for(int i = 1;i<=21;i++){
        for(int j = 1;j+(1<<i)-1<=N;j++){
            a[j][i] = max(a[j][i-1],a[j+(1<<(i-1))][i-1]);
        }
    }
    int t1,t2;
    for(int i = 1;i<=M;i++){
        t1=qr();t2=qr();
        int k = log2(t2-t1+1);
        printf("%d
",max(a[t1][k],a[t2-(1<<k)+1][k]));
    }
    return 0;
}