$NOIP2002$ 题解报告

目录

$Luogu\ P1031$ 均分纸牌$(\ √\ )$

$Luogu\ P1032$ 字串变换$(\ √\ )$

$Luogu\ P1033$ 自由落体$(\ √\ )$

$Luogu\ P1034$ 矩形覆盖$(\ √\ )$

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$Luogu\ P1031$ 均分纸牌

题目传送门

先算出平均值，然后处理出每堆纸牌与平均值的关系，多了则为$+$，少了为$-$。然后每次把当前这一堆的牌改变，加到下一堆牌里，这样一定是最优的

#include<bits/stdc++.h>
using namespace std;
int a[102];
int main(){
    int n,i,ave=0,step=0;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];ave+=a[i];
    }
    ave/=n;
    for(i=1;i<=n;i++) a[i]-=ave;
    i=1;
    int j=n;
    while(a[i]==0&&i<n) i++;
    while(a[j]==0&&j>1) j--;
    while(i<j)
     {
         a[i+1]+=a[i];
         a[i]=0;
         step++;i++;
         while(a[i]==0&&i<j) i++;
     }
    cout<<step<<endl;
    return 0;
}

 

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$Luogu\ P1032$ 字串变换

题目传送门

直接$bfs$，枚举每个转换方式能否匹配，用$map$判重

#include<bits/stdc++.h>
#define ri register int
#define ll long long
#define rl register ll
#define go(i,a,b) for(ri i=a;i<=b;i++)
#define back(i,a,b) for(ri i=a;i>=b;i--)
#define g() getchar()
#define il inline
#define pf printf
#define sf scanf
#define ull unsigned ll
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
il int fr(){
    ri w=0,q=1;char ch=g();
    while(ch<'0'||ch>'9'){if(ch=='-')q=-1;ch=g();}
    while(ch>='0'&&ch<='9')w=(w<<1)+(w<<3)+ch-'0',ch=g();
    return w*q;
}
const int N=42;
string A,B,a[10],b[10];
struct node{string s;int t;}p;
queue<node> q;
map<string,int> vis;
int n=1,lena[10],lenb[10];
string change(const string &x,ri i,ri j){
    string as="";
    if(j+lena[i]>x.length())return as;
    go(k,0,lena[i]-1)if(x[j+k]!=a[i][k])return as;
    as=x.substr(0,j);
    as+=b[i];as+=x.substr(j+lena[i]);
    return as;
}
int main(){
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    cin>>A>>B;q.push((node){A,0});vis[A]=1;
    while(cin>>a[n]>>b[n]){lena[n]=a[n].length();lenb[n]=b[n].length();n++;}n--;
    while(!q.empty()){
        p=q.front();q.pop();
        if(p.t>10)break;
        if(p.s==B){pf("%d\n",p.t);return 0;}
        ri len=p.s.length();
        go(i,1,n)go(j,0,len-1){
            string chg=change(p.s,i,j);
            if(chg!=""){
                node nw;nw.s=chg;nw.t=p.t+1;
                if(!vis.count(nw.s))q.push(nw),vis[nw.s]=1;
            }
        }
    }
    puts("NO ANSWER!");
    return 0;
}

 

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$Luogu\ P1033$ 自由落体

题目传送门

算出小车能接到的小球的位置范围，直接统计个数即可

#include<bits/stdc++.h>
using namespace std;
float H,S,V,L,K,n;
int ans=0;
int main(){
    scanf("%f%f%f%f%f%f",&H,&S,&V,&L,&K,&n);
    float t1,t2,l,r;
    t1=sqrt((H-K)/5);
    t2=sqrt(H/5);
    l=S-V*t2-0.0001;
    r=S+L-V*t1+0.0001;
    for(int i=max(0,int(ceil(l)));i<=min(n-1,r);i++)
      ans++;
    printf("%d",ans);
    return 0;
}

 

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$Luogu\ P1034$ 矩形覆盖

题目传送门

$dfs$大法好，每次枚举当前这个点加到哪个矩阵里去，剪枝就是最优性剪枝$+$判断当前情况下矩形是否有重叠

#include<bits/stdc++.h>
#define ri register int
#define ll long long
#define rl register ll
#define go(i,a,b) for(ri i=a;i<=b;i++)
#define back(i,a,b) for(ri i=a;i>=b;i--)
#define g() getchar()
#define il inline
#define pf printf
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
il int fr(){
    ri w=0,q=1;char ch=g();
    while(ch<'0'||ch>'9'){if(ch=='-')q=-1;ch=g();}
    while(ch>='0'&&ch<='9')w=(w<<1)+(w<<3)+ch-'0',ch=g();
    return w*q;
}
const int N=52;
int n,k,R,ans=1e9+7,a[N],b[N];
struct node{int x,y;}p[N];
struct square{int x1,y1,x2,y2,s;}q[5],Q[5];
il bool cmp(node a,node b){return a.x==b.x?a.y<b.y:a.x<b.x;}
il bool CMP(square a,square b){return a.x1<b.x1;}
il bool check(ri num){
    go(i,1,num)Q[i]=q[i];
    sort(Q+1,Q+1+num,CMP);
    go(i,1,num-1)
        if(Q[i+1].x1<=Q[i].x2&&((Q[i+1].y1<=Q[i].y2&&Q[i+1].y1>=Q[i].y1)||(Q[i+1].y2>=Q[i].y1&&Q[i+1].y1<=Q[i].y1)))return 0;
    return 1;
}
il void work(ri id,ri num,ri sum){
    if(sum>=ans||!check(num))return;
    //cout<<"id="<<id<<" num="<<num<<" sum="<<sum<<endl;
    if(id>n){if(num==k)ans=sum;return;}
    ri x=p[id].x,y=p[id].y;
    back(i,num,1){
        square bf=q[i];
        if(y<q[i].y1)q[i].y1=y;
        if(y>q[i].y2)q[i].y2=y;
        if(x>q[i].x2)q[i].x2=x;
        ri add=(q[i].x2-q[i].x1)*(q[i].y2-q[i].y1)-q[i].s;
        q[i].s=(q[i].x2-q[i].x1)*(q[i].y2-q[i].y1);
        work(id+1,num,sum+add);q[i]=bf;
    }
    if(num<k){
        q[num+1]=(square){x,y,x,y,0};
        work(id+1,num+1,sum);
    }
    return;
}
int main(){
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    n=fr();k=fr();
    go(i,1,n)p[i]=(node){fr(),fr()};
    sort(p+1,p+1+n,cmp);p[0].x=-1;
    work(1,0,0);
    pf("%d\n",ans);
    return 0;
}