平时十二测

题解：

第一题：简单模拟：

#include<bits/stdc++.h>
using namespace std;

const int M = 500000;// up;
char s[M], ans[M];

int main(){
    freopen("expression.in","r",stdin);
    freopen("expression.out","w",stdout);
    int tot = -1;
    scanf("%s", s);
    int len = strlen(s);
    for(int i = 0; i < len;){
        if(s[i] == ‘+‘){
            ans[++tot]=‘+‘;i++;
            while(i<len&&s[i]>=‘0‘&&s[i]<=‘9‘)ans[++tot]=s[i++];
        }
        if(i==len)break;
        if(s[i] == ‘-‘){
            ans[++tot]=‘-‘;
            i++;
            int t=i;
            if(s[i]==‘0‘){ans[++tot]=‘0‘;i++;continue;}
            if(s[i]!=‘0‘){
                ans[++tot]=s[i++];
                if(i==len)break;
                int t=i;
                while(i<len&&s[i]==‘0‘){
                    ans[++tot]=‘+‘;
                    ans[++tot]=‘0‘;
                    i++;
                }
                if(i==len)break;
                if(s[i]==‘+‘||s[i]==‘-‘)continue;
                else {
                    ans[++tot]=‘+‘;
                    while(i<len&&s[i]>=‘0‘&&s[i]<=‘9‘){
                        ans[++tot]=s[i++];
                    }
                }
                if(i==len)break;
            }

        }
        if(s[i]<=‘9‘&&s[i]>=‘0‘&&i==0){
            while(i<len&&s[i]<=‘9‘&&s[i]>=‘0‘)
                ans[++tot]=s[i++];
        }
    }
    printf("%s", ans);
}

第二题：

#include<bits/stdc++.h>
using namespace std;

const int M = 400005, N = 100005;
int h[N], fa[N], tot = 1, dep[N], cnto, cnte, even[N], odd[N];
bool vis[M];
struct edge{int v, nxt;}G[M];
struct node{int u, v, id;}g[M];
void add(int u, int v){
    G[++tot].v =v, G[tot].nxt = h[u], h[u] = tot;
}
int dfs1(int u, int f){
    dep[u] = dep[f] + 1;
    for(int i = h[u]; i; i = G[i].nxt){
        if(vis[i])continue;
        int v = G[i].v;
        vis[i] = vis[i^1] = 1;
        if(dep[v]){
            if((dep[v]&1) == (dep[u]&1)){
                odd[u]++,odd[v]--,cnto++;
            }
            else even[u]++, even[v]--,cnte++;
        }
        else {
            fa[v] = i;
            dfs1(v, u);
        }
    }

}
void dfs2(int u){

    for(int i = h[u]; i; i = G[i].nxt){
        if(i == fa[G[i].v]){
            int v=G[i].v;
            dfs2(v);
            odd[u] += odd[v], even[u] += even[v];
        }
    }
}
int read(){
    int x = 0; int f = 1; char c = getchar();
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c<=‘9‘&&c>=‘0‘){x=x*10+c-‘0‘;c=getchar();}
    return x*=f;
}

int main(){

    freopen("voltage.in","r",stdin);
    freopen("voltage.out","w",stdout);
        int n = read(), m = read();
        int ans = 0;
        for(int i = 1; i <= m; i++){
            int u = read(), v = read();
            add(u, v);
            add(v, u);
        }
        dfs1(1, 0);
        dfs2(1);
        for(int i = 2; i <= n; i++)
            if(!even[i] && odd[i]==cnto) ans++;
        if(cnto == 1) ans++;
        printf("%d\n", ans);
}

第三题：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std ; 

int L , N , jl[2005] , jr[2005] , rg[2005] , cok[2] , nw , ans , nd1 , nd2 ;
char ty[2005] ;  

void insert ( int l , int r )
{
    while ( rg[l] )
    {
        if ( rg[l] > r ) { int ll = l ; l = r + 1 ; swap ( rg[ll] , r ) ; }
        else if ( rg[l] < r ) l = rg[l] + 1 ;
        if ( rg[l] == r || l >= L ) return ;
    }
    rg[l] = r ;
}

int zh ( char x ) { if ( x == ‘(‘ ) return 0 ; return 1 ; }

void gao ( int l , int r )
{
    int mx = 0 ;
    while ( nw <= r )
    {
        if ( ty[nw] == ‘(‘ ) mx ++ ;
        else if ( ty[nw] == ‘)‘ )
        {
            if ( mx ) mx-- ;
            else { nd1++ ; mx++ ; }
        }
        nw++ ;
        if ( rg[nw] )
        {
            if ( rg[nw] > r && nw <= r )
            {
                insert ( r + 1 , rg[nw] ) ;
                insert ( nw , r ) ;
            }
            gao ( nw , rg[nw] ) ;
        }
    }
    if ( mx % 2 ) ans = -1e9 ;
    nd2 += mx / 2 ;
}

int main ( )
{
    freopen ( "parentheses.in" , "r" , stdin ) ;
    freopen ( "parentheses.out" , "w" , stdout ) ;
    scanf ( "%s" , &ty ) ;
    L = strlen ( ty ) ;
    scanf ( "%d" , &N ) ;
    for ( int i = 1 ; i <= N ; i++ ) scanf ( "%d" , &jl[i] ) ;
    for ( int i = 1 ; i <= N ; i++ ) scanf ( "%d" , &jr[i] ) ;
    for ( int i = 1 ; i <= N ; i++ )
    {
        if ( ( jr[i] - jl[i] + 1 ) % 2 ) { printf ( "-1\n" ) ; return 0 ; }
        insert ( jl[i] , jr[i] ) ;
    }
    while ( nw < L )
    {
        if ( !rg[nw] ) cok[zh ( ty[nw++] )]++ ;
        else gao ( nw , rg[nw] ) ;
    }
    while ( nd1 && nd2 ) ans ++ , nd1 -- , nd2 -- ;
    if ( cok[0] < nd1 || cok[1] < nd2 ) ans = -1e9 ;
    ans += ( nd1 + nd2 ) ;
    ans < 0 ? printf ( "-1\n" ) : printf ( "%d\n" , ans ) ;
}

原文地址：https://www.cnblogs.com/EdSheeran/p/9822131.html