20180705 考试记录

Posted 2020-12-02 menteur-hxy

T1 货物运输弱化版

题解：

倒着跑最短路就行
没仔细看题凉凉
code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
using namespace std;

int rd() {
    int x=0,f=1; char c=getchar();
    while(!isdigit(c)) {if(c==‘-‘)f=-f; c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
    return x*f;
}

const int N=5010;
int n,cnt,tot;
int nxt[N],head[N],to[N],qu[N<<3],dp[N],vis[N];

void bfs(int x) {//SPFA
    int h=0,t=1;qu[h]=x;
    while(h<=t) {
        int u=qu[h++],ret=0;
        vis[u]=0;
        if(u<27) {
            ret=dp[u]/19;
            if(ret*19!=dp[u]) ret++;
        } else ret=1;
        for(int i=head[u];i;i=nxt[i]) {
            int v=to[i];
            if(dp[v]>dp[u]+ret) {
                dp[v]=dp[u]+ret;
                if(!vis[dp[v]]) qu[t++]=v,vis[v]++;
            }
        }
    }
}

#define add(a,b) nxt[++cnt]=head[a],to[cnt]=b,head[a]=cnt
#define insert(a,b) add(a,b),add(b,a)
int main() {
    freopen("toll.in","r",stdin);
    freopen("toll.out","w",stdout);
    while(scanf("%d",&n),n!=-1) {
        cnt=0;tot++;
        memset(vis,0,sizeof(vis));
        memset(dp,0x3f,sizeof(dp));
        memset(head,0,sizeof(head));
        char a,b;
        F(i,1,n) {
            scanf(" %c %c",&a,&b);
            insert(a-‘A‘+1,b-‘A‘+1);
        }
        getchar();
        int ret; scanf("%d ",&ret);
        scanf("%c %c",&a,&b);
        int T=a-‘A‘+1,S=b-‘A‘+1;
        dp[S]=ret; bfs(S);
        printf("Case %d: %d
",tot,dp[T]);//考试时直接输出dp[T]QAQ凉凉 
    }
    return 0;
}

T2 冒泡排序

求没有逆序对的最长序列——不就是LIS么？ ——高 Orz
code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int rd() {
    int x=0,f=1; char c=getchar();
    while(!isdigit(c)) {if(c==‘-‘)f=-f; c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
    return x*f;
}
const int N=100010;
int n,top,da[N],dp[N];
int main() {
    freopen("sort.in","r",stdin);
    freopen("sort.out","w",stdout);
    n=rd();
    for(int i=1;i<=n;i++) da[i]=rd();
    for(int i=1;i<=n;i++) if(dp[top]<da[i]) dp[++top]=da[i];
    else dp[lower_bound(dp+1,dp+1+top,da[i])-dp]=da[i];
    printf("%d",top);
    return 0;
}

T3 room

比较水的状压
dp[i][j] i表示状态 j表示不同颜色钥匙数量
code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
using namespace std;

int rd() {
    int x=0,f=1; char c=getchar();
    while(!isdigit(c)) {if(c==‘-‘)f=-f; c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
    return x*f;
}

int n,re,gr,wi,ans,mx;
int rek[15],grk[15],wik[15],red[15],grd[15];
int dp[20000][3];

bool jud(int a,int b) {
    int sum=0;
    int ret=dp[a-(1<<b)][0]-red[b+1]; if(ret<0) sum+=ret;
    ret=dp[a-(1<<b)][1]-grd[b+1]; if(ret<0) sum+=ret;
    return sum+dp[a-(1<<b)][2]>=0;
}

int main() {
    freopen("room.in","r",stdin);
    freopen("room.out","w",stdout);
    n=rd();
    F(i,1,n) red[i]=rd();
    F(i,1,n) grd[i]=rd();
    F(i,1,n) rek[i]=rd();
    F(i,1,n) grk[i]=rd();
    F(i,1,n) wik[i]=rd();
    dp[0][0]=rd(),dp[0][1]=rd(),dp[0][2]=rd();
    ans=dp[0][0]+dp[0][1]+dp[0][2];
    for(int i=0;i<(1<<n);i++) {
        for(int j=0;(i>>j);j++) 
            if(((i>>j)&1)&&jud(i,j)) {
                dp[i][2]=dp[i-(1<<j)][2]+wik[j+1];
                
                if(dp[i-(1<<j)][0]-red[j+1]<0) {
                    dp[i][2]+=dp[i-(1<<j)][0]-red[j+1];
                    dp[i][0]=rek[j+1];
                } else dp[i][0]=dp[i-(1<<j)][0]-red[j+1]+rek[j+1];
                
                if(dp[i-(1<<j)][1]-grd[j+1]<0) {
                    dp[i][2]+=dp[i-(1<<j)][1]-grd[j+1];
                    dp[i][1]=grk[j+1];
                } else dp[i][1]=dp[i-(1<<j)][1]-grd[j+1]+grk[j+1];
                ans=max(ans,dp[i][0]+dp[i][1]+dp[i][2]);
            }
    }
    printf("%d",ans);
    return 0;
}