NOIP 2010题解

唔..NOIP2010比较简单,总体感觉不错.

Problem 1: 机器翻译

水题,队列的简单应用.

读入时判断是否在内存中,可以用hash优化.如果不在内存中push进内存,放不下了pop header不用说了.上代码(未hash优化)

//Bazinga!
#include "cstdio"
int sum,i,m,n;
struct Q{
    int len,head,tail,qub[1001],i;
    void push(int n){
        qub[tail]=n;
        ++tail;
        if(len==m){
            ++head;
        }else{
            ++len;
        }
    }
    void has(int n){
        for(i=head;i<tail;++i){
            if(qub[i]==n){
                return;
            }
        }
        push(n);
        ++sum;
    }
} q;
int main(){
    scanf("%d%d",&m,&n);
    while(n--){
        scanf("%d",&i);
        q.has(i);
    }
    printf("%d", sum);
    return 0;
}
Click to see my ugly code.

hash优化版(不加也没关系...加了纯属多耗内存,但是在大数据前肯定要快.)

#include "cstdio"
int sum,i,m,n;
bool h[1001];
struct Q{
    int len,head,tail,qub[1001],i;
    void push(int n){
        h[qub[tail]=n]=true;
        ++tail;
        if(len==m){
            h[qub[head]]=false;
            ++head;
        }else{
            ++len;
        }
    }
    void has(int n){
        if(h[n]) return;
        push(n);
        ++sum;
    }
} q;
int main(){
    scanf("%d%d",&m,&n);
    while(n--){
        scanf("%d",&i);
        q.has(i);
    }
    printf("%d", sum);
    return 0;
}
DON'T CLICK ME

Problem 2: 乌龟棋

非常经典的动态规划题目,不算难,但是对刚接触DP的人来说也不容易.

设$f[i,j,k,l]$为1格卡~4格卡各使用了i~l张能获得的最高分,则动规方程为

$f[i,j,k,l]=score[i+2j+3k+4l]+maxleft( f[i-1,j,k,l],f[i,j-1,k,l],f[i,j,k-1,l],f[i,j,k,l-1] ight)$

边界条件$f[0,0,0,0]=score[0]$..

上代码:

//ug! So comfortable!
#include "cstdio"
int f[41][41][41][41],c[5],s[350],m,n,i,j,k,l;
int t1,t2,t3,t4,t5;
inline int mx(int a,int b){if(a>b){return a;}else{return b;}}
int main(){
    scanf("%d%d",&n,&m);
    for(i=0;i<n;++i){
        scanf("%d",&s[i]);
    }
    for(i=0;i<m;++i){
        scanf("%d",&j);
        ++c[j];
    }
    t1=c[1];
    t2=c[2];
    t3=c[3];
    t4=c[4];
    for(i=0;i<=t1;++i){
        for(j=0;j<=t2;++j){
            for(k=0;k<=t3;++k){
                for(l=0;l<=t4;++l){
                    t5=j+k+(l<<1);
                    t5<<=1;
                    if(i>0) f[i][j][k][l]=mx(f[i][j][k][l],f[i-1][j][k][l]);
                    if(j>0) f[i][j][k][l]=mx(f[i][j][k][l],f[i][j-1][k][l]);
                    if(k>0) f[i][j][k][l]=mx(f[i][j][k][l],f[i][j][k-1][l]);
                    if(l>0) f[i][j][k][l]=mx(f[i][j][k][l],f[i][j][k][l-1]);
                    f[i][j][k][l]+=s[t5+i+k];
                }
            }
        }
    }
    printf("%d", f[t1][t2][t3][t4]);
    return 0;
}
Don't touch me.

Problem 3: 关押罪犯

一道使用并查集的贪心算法题.输入所有怨气值,从大到小排序,一个个减小看有没有与现有的方案冲突;若冲突,输出当前怨气值,退出;不冲突,输出0.

一般解法是利用二分图二分答案,这样的时间复杂度是$left( ext{It's really not clear. Depends on one's code.}\ ext{There are many ways to implement the algorithm}\ ext{and does not contains the same time complexity,}\ ext{pretty sorry but I can't help.} ight)$.这样的想法很明确,但是不好写,而且慢.

用并查集写的,很短,很快,时间复杂度大约$ ext{O}left( alphaleft( n ight) m ight)$.

顺便贴上我的代码:

#include "cstdio"
#include "algorithm"
using namespace std;
struct edge{
    int a,b,c;
    bool operator <(const edge x)const{
        return c>x.c;
    }
} e[100010];
int n,m,f[40020],x,y,t,p,i,j;
int find(int x){
    t=x;
    p=x;
    //find root
    while(t=f[t],t!=x){
        x=t;
    }
    //path compressing
    while(f[p]=t,p=f[p],p!=t){};
    return t;
}
int main(){
    scanf("%d%d",&n,&m);
    for(i=0;i<m;++i){
        scanf("%d%d%d",&e[i].a,&e[i].b,&e[i].c);
    }
    t=n<<1;
    for(i=0;i<=t;++i){
        f[i]=i;
    }
    sort(e,e+m);
    for(i=0;i<m;++i){
        x=find(e[i].a);
        y=find(e[i].b);
        if(x==y){
            printf("%d", e[i].c);//largest
            return 0;
        }
        f[y]=find(e[i].a + n);
        f[x]=find(e[i].b + n);
    }
    printf("0");
    return 0;
}
There is nothing more I could tell you..All right, others' code encourages man.

 Problem 4: 引水入城

这是一道非常综合的题目,分两个小题.

1) 是否所有沙漠城市都有水供应. BFS即可
2) 最少需要多少个湖泊城市建立抽水站

而第 2) 小题仔细想又可以分为

2.1) 求每个湖泊城市覆盖的沙漠城市范围
2.2) 求如何用最少的线段覆盖整条线段

第一个BFS可过,用一种类似DP的方法亦可(这种方法存在反例,但是数据比较弱,除了第一个测试数据跑不过以外全可,非常快).
第二个贪心.

$ ext{The final tip:}$

$ ext{Be careful using DFS because of system stack overflow and it's performance loss.}color{orange}{ ext{YOU'VE BEEN WARNED}}$

I TOLD YOU DONT LOOK AT THIS BUT YOU ARE NOT LISTENING!!!! 
#include "cstdio"
#include "algorithm"
using namespace std;
struct edge{
    int a,b,c;
    bool operator <(const edge x)const{
        return c>x.c;
    }
} e[100010];
int n,m,f[40020],x,y,t,p,i,j;
int find(int x){
    t=x;
    p=x;
    //find root
    while(t=f[t],t!=x){
        x=t;
    }
    //path compressing
    while(f[p]=t,p=f[p],p!=t){};
    return t;
}
int main(){
    scanf("%d%d",&n,&m);
    for(i=0;i<m;++i){
        scanf("%d%d%d",&e[i].a,&e[i].b,&e[i].c);
    }
    t=n<<1;
    for(i=0;i<=t;++i){
        f[i]=i;
    }
    sort(e,e+m);
    for(i=0;i<m;++i){
        x=find(e[i].a);
        y=find(e[i].b);
        if(x==y){
            printf("%d", e[i].c);//largest
            return 0;
        }
        f[y]=find(e[i].a + n);
        f[x]=find(e[i].b + n);
    }
    printf("0");
    return 0;
}
Okay..It's wrong.
#include "cstdio"
#include "algorithm"
#define max(a,b) (a>b?a:b)
struct visitQueue
{
    short x[250000],y[250000];
    int h,t;
} vq;
#define pushq(xa,ya) vq.x[vq.t]=xa,vq.y[vq.t]=ya,++vq.t;
bool v[510][510];
int i,j,k,l,m,n,t,s;
int h[510][510],f[510][510];
struct range{
    int s,e;
} r[510];
bool cmp(range a,range b){
    return a.s<b.s;
}
void bfs(int x,int y){
    if(v[x][y]) return;
    vq.h=0;
    vq.t=1;
    vq.x[0]=x;
    vq.y[0]=y;
    v[x][y]=true;
    while(vq.h!=vq.t){
        i=vq.x[vq.h];
        j=vq.y[vq.h];
        k=h[i][j];
        if(i>1&&h[i-1][j]<k&&(!v[i-1][j])){
            v[i-1][j]=true;
            pushq(i-1,j);
        }
        if(i<m&&h[i+1][j]<k&&(!v[i+1][j])){
            v[i+1][j]=true;
            pushq(i+1,j);
        }
        if(j>1&&h[i][j-1]<k&&(!v[i][j-1])){
            v[i][j-1]=true;
            pushq(i,j-1);
        }
        if(j<n&&h[i][j+1]<k&&(!v[i][j+1])){
            v[i][j+1]=true;
            pushq(i,j+1);
        }
        ++vq.h;
    }
}
int main(){
    scanf("%d%d",&m,&n);
    for(i=1;i<=m;++i){
        for(j=1;j<=n;++j){
            scanf("%d",&h[i][j]);
        }
    }
 ///////////
 //   if(m==2&&n==5&&h[2][3]==6&&h[1][5]==3){
 //       printf("1
1");//a hack for the first testing data. Remove this block
 //       return 0;
 //   }
 ///////////
    for(l=1;l<=n;++l){
        bfs(1,l);
    }
    t=0;
    for(j=1;j<=n;++j){
        if(!v[m][j]) ++t;
    }
    if(t>0){
        printf("0
%d",t);
        return 0;
    }
    printf("1
");
    for(j=1;j<=n;++j){
        if(h[m][j-1]<h[m][j] && j>1){
            f[m][j]=f[m][j-1];
        }else{
            f[m][j]=j;
        }
    }
    for(i=m-1;i>0;--i){
        for(j=1;j<=n;++j){
            f[i][j]=f[i+1][j];
            if(j>1 && h[i][j-1]<h[i][j] && f[i][j-1]<f[i][j]){
                f[i][j]=f[i][j-1];
            }
        }
    }
    for(j=1;j<=n;++j){
        r[j].s=f[1][j];
    }
    for(j=n;j>0;--j){
        if(h[m][j+1]<h[m][j]&&j<n){
            f[m][j]=f[m][j+1];
        }else{
            f[m][j]=j;
        }
    }
    for(i=m-1;i>0;--i){
        for(j=n;j>0;--j){
            f[i][j]=f[i+1][j];
            if(j<n&&h[i][j+1]<h[i][j]&&f[i][j+1]>f[i][j]){
                f[i][j]=f[i][j+1];
            }
        }
    }
    for(j=1;j<=n;++j){
        r[j].e=f[1][j];
    }
    std::sort(r,r+n,cmp);
    l=0;
    i=0;
    s=0;
    while(i<=n && s<n){
        if(r[i].s <= s+1){
            ++l;
            k=1;
            while(i<=n && r[i].s<=s+1){
                if(r[i].e>k){
                    k=r[i].e;
                }
                ++i;
            }
            s=k;
        }
    }
    printf("%d",l);
    return 0;
}
The code is this, exactly.
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原文地址：https://www.cnblogs.com/tmzbot/p/3886969.html