算法竞赛入门经典第六章总结

并行程序模拟：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1010;

deque<int>wait;
queue<int>block;

int line,quan;
int pgline[100];
char pgcmd[maxn][10];
int val[26];

int t[5],locked;

void run(int num);

int main(void)
{
    memset(pgline,0,sizeof(pgline));
    memset(pgcmd,0,sizeof(pgcmd));
    memset(val,0,sizeof(val));
    int n; cin >> n;
    for(int i=0;i<5;i++) cin >> t[i];
    cin >> quan;
    
    for(int i=0;i<n;i++)
    {
        gets(pgcmd[line++]);
        pgline[i]=line-1;
        while(pgcmd[line-1][2]!='d') gets(pgcmd[line++]);
        wait.push_back(i);
    }
    
    locked=0;
    while(!wait.empty())
    {
        int pick = wait.front();
        wait.pop_front();
        run(pick);
    }
    return 0;
} 

void run(int num)
{
    int q=quan;
    while(q>0)
    {
        char *str=pgcmd[pgline[num]];
        switch(str[2])
        {
            case '=':
                val[str[0]-'a']=isdigit(str[5])?(str[4]-'0')*10+(str[5]-'0'):str[4]-'0';
                q-=t[0];
                break;
            case 'i':
                printf("%d : %d\n",num+1,val[str[6]-'0']);
                q-=t[1];
                break; 
            case 'c':
                if(locked){
                    block.push(num);
                    return ;
                }
                locked=1;
                q-=t[2];
                break;
            case 'l':
                locked=0;
                if(!block.empty())
                {
                    wait.push_front(block.front());
                    block.pop();
                }
                q-=t[3];
                break;
            case 'd':
                return ;
        }
        pgline[num]++;
    }
}

 

铁轨：

#include<stack>
#include<iostream>
#include<cstdio>
using namespace std;

const int maxn = 1010;

stack<int>s;
int ans[maxn];

int main(void)
{
    int n;
    cin >> n;
    for(int i=1;i<=n;i++) cin >> ans[i];
    int A=1,B=1,ok=1;
    while(B<=n)
    {
        if(A==ans[B]) {
            A++,B++;
        }
        else if(!s.empty() && s.top()==ans[B])
        {
            B++;
            s.pop(); 
        }
        else if(A<=n)
        {
            s.push(A++);    
        } 
        else {
            ok = 0;
            break;
        }
    }
    //cout << ok?"YES":"NO" << endl;
    printf("%s\n",ok?"YES":"NO");
    return 0;
}

 

矩阵链乘：

#include<iostream>
#include<stack>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;

struct Matrix{
    int a,b;
    Matrix(int a=0,int b=0):a(a),b(b){}
    
}m[26];

stack<Matrix>s;

int main(void)
{
    int n;
    cin >> n;
    for(int i=0;i<n;i++) {
        string name;
        cin >> name;
        int k=name[0]-'A';
        cin >> m[k].a >> m[k].b; 
    }    
    string expr;
    while(cin >> expr)
    {
        int len=expr.length();
        bool error = false;
        int ans = 0;
        for(int i=0;i<len;i++)
        {
            if(isalpha(expr[i])) s.push(m[expr[i]-'A']);
            else if(expr[i]==')') {
                Matrix m2=s.top(); s.pop();
                Matrix m1=s.top(); s.pop();
                if(m1.b!=m2.a){
                    error=true;
                    break;
                }
                ans+=m1.a*m1.b*m2.b;
                s.push(Matrix(m1.a,m2.b));
            }
        }
        if(error) printf("error\n"); else printf("%d\n",ans);
    }
}