tyvj NOIP2017金秋冲刺训练营杯联赛模拟大奖赛第一轮Day2题解

　　上星期打的...题有点水，好多人都AK了

　　T1排个序贪心就好了

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=500010,inf=1e9;
struct poi{int e,s;}a[maxn];
int n,m,s,k,x,y,cnt;
int cp[maxn];
ll ans;
void read(int &k)
{
    int f=1;k=0;char c=getchar();
    while(c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1),c=getchar();
    while(c<=‘9‘&&c>=‘0‘)k=k*10+c-‘0‘,c=getchar();
    k*=f;
}
inline bool cmp(poi a,poi b){return a.e==b.e?a.s>b.s:a.e<b.e;}
inline bool cmp2(poi a,poi b){return a.s>b.s;}
int main()
{
    freopen("express.in","r",stdin);
    freopen("express.out","w",stdout);
    read(n);read(m);read(s);read(k);
    for(int i=1;i<=m;i++)
    {
        read(x);read(y);
        if(y<=2)continue;
        a[++cnt].e=x;a[cnt].s=y-2;
    }
    sort(a+1,a+1+cnt,cmp);
    for(int i=1;i<=cnt;i++)if(a[i].e!=a[i-1].e)cp[a[i].e]=i;
    for(int i=1;i<=n;i++)
    {
        if(!cp[i])return puts("-23333333"),0;
        int kk=k;
        for(int j=cp[i];j<=cnt&&a[j].e==i;j++)
        if(kk>0)
        {
            if(!s)return puts("-23333333"),0;
            s--;kk-=a[j].s;ans+=a[j].s;a[j].s=0;
        }
        if(kk>0)return puts("-23333333"),0;
    }
    if(s)
    {
        sort(a+1,a+1+cnt,cmp2);
        for(int i=1;i<=s;i++)
        ans+=a[i].s;
    }
    printf("%lld\n",ans);
    return 0;
}

　　T2我写的方法就太鶸了，所以只说某个很妙的做法（%%%腾腾太强辣！）

　　用一个指针扫，扫到某个括号就跳到对应的括号然后反向，遇到字符输出，模拟一下就知道是非常正确的

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=500010,inf=1e9;
char s[maxn];
int n,top;
int st[maxn],op[maxn];
void read(int &k)
{
    int f=1;k=0;char c=getchar();
    while(c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1),c=getchar();
    while(c<=‘9‘&&c>=‘0‘)k=k*10+c-‘0‘,c=getchar();
    k*=f;
}
int main()
{
    freopen("unknown.in","r",stdin);
    freopen("unknown.out","w",stdout);
    scanf("%s",s+1);n=strlen(s+1);
    for(int i=1;i<=n;i++)
    {
        if(s[i]==‘(‘)st[++top]=i;
        if(s[i]==‘)‘)op[i]=st[top],op[st[top--]]=i;
    }
    for(int i=1,st=1;i<=n;i+=st)
    if(s[i]==‘(‘||s[i]==‘)‘)i=op[i],st=-st;
    else printf("%c",s[i]);
    return 0;
}

　　T3状压，枚举子集的子集是3^n的，于是可以过

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=20;
int n,m,K,x,y;
int f[1<<maxn];
bool v[maxn][maxn],mp[1<<maxn];
void read(int &k)
{
    int f=1;k=0;char c=getchar();
    while(c<‘0‘||c>‘9‘)c==‘-‘&&(f=-1),c=getchar();
    while(c<=‘9‘&&c>=‘0‘)k=k*10+c-‘0‘,c=getchar();
    k*=f;
}
int main()
{
    freopen("prison.in","r",stdin);
    freopen("prison.out","w",stdout);
    read(n);read(m);read(K);
    for(int i=1;i<=m;i++)read(x),read(y),v[x][y]=v[y][x]=1;
    int state=(1<<n)-1;
    for(int i=0;i<=state;i++)
    {
        int cnt=0;
        for(int j=1;j<=n;j++)
        if(i&(1<<(j-1)))
        for(int k=1;k<j;k++)
        if(i&(1<<(k-1)))
        cnt+=v[j][k];
        if(cnt<=K)mp[i]=1;
    }
    memset(f,32,sizeof(f));f[0]=0;
    for(int i=0;i<=state;i++)
    {
        int S=i^state;
        for(int j=S;j;j=(j-1)&S)
        if(mp[j])f[i|j]=min(f[i|j],f[i]+1);
    }
    printf("%d\n",f[state]);
    return 0;
}