题解：

第一题：简单模拟：

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

 

第二题：

 

 

#include
      
       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 
      
       #include 
       
        #include 
        
         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 ) ;     }