#include<iostream>
#include<cstring>
#include<math.h>
#include<cstdio>
#include<algorithm>
#define N 6005
#define INF 0x3f3f3f3f
#include<map>
#include<string>
#include<queue>
#include<vector>
#include<set>
#define mod 1000000007
double pi=acos(-1.0);
typedef long long ll;
using namespace std;
typedef pair<ll,int
 
> pr;
priority_queue<pr,vector<pr>,greater<pr> > q[4];
ll p[200050],a[200050],b[200050];
int vis[200050];
int m,c,n;
int main(){
    memset(vis,0,sizeof(vis));
    cin>>n;
    for(int i=0;i<n;i++)
    {
        scanf("%I64d",&p[i]);
    }
    for(int i=0;i<n;i++)
    {
        scanf("%d
 
",&a[i]);
    }
    for(int i=0;i<n;i++)
    {
        scanf("%d",&b[i]);
    }
    for(int i=0;i<n;i++)
    {
        q[a[i]].push(make_pair(p[i],i));
        q[b[i]].push(make_pair(p[i],i));
    }
    cin>>m;
    for(int i=0;i<m;i++)
    {
        scanf("%d",&c);
        int
 
 flag=0;
        while(!q[c].empty())
        {
            if(vis[q[c].top().second]) q[c].pop();
            else
            {
                flag=1;
                break;
            }
        }
        if(!flag)
        {
            printf("-1 ");
            continue;
        }
        pr temp=q[c].top();
        printf("%I64d ",temp.first);
        q[c].pop();
        vis[temp.second]=1;
    }
    return 0;
}

2、cf 3D D. Least Cost Bracket Sequence
把所有的‘？’都換成右括號，放進佇列，當不滿足對稱時選cost最小的就可以了
程式碼：

#include<iostream>
#include<cstring>
#include<math.h>
#include<cstdio>
#include<algorithm>
#define N 6005
#define INF 0x3f3f3f3f
#include<map>
#include<string>
#include<queue>
#include<vector>
#include<set>
#define mod 1000000007
double pi=acos(-1.0);
typedef long long ll;
using namespace std;
char s[50050];
int a,b;
typedef pair<int,int> p;
int main(){
    scanf("%s",s);
    int m=0,zuo=0;
    ll ans=0;
    int len=strlen(s);
    if(len&1||s[0]==')')
    {
        printf("-1\n");
        return 0;
    }
    priority_queue<p> q;
    for(int i=0;i<len;i++)
    {
        if(s[i]=='(') zuo++;
        else if(s[i]==')') zuo--;
        else
        {
            cin>>a>>b;
            ans+=b;
            s[i]=')';
            q.push(make_pair(b-a,i));
            zuo--;
        }
        if(zuo<0)
        {
            if(q.empty())
            {
                printf("-1\n");
                return 0;
            }
            p temp=q.top();
            q.pop();
            zuo+=2;
            ans-=temp.first;
            s[temp.second]='(';
        }
    }
    if(zuo>0)
    {
        printf("-1\n");
        return 0;
    }
    else printf("%I64d\n%s\n",ans,s);
    return 0;
}

3、ZOJ 3699
經典的加油問題
程式碼：

#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <queue>  
#include <iostream>  
#include <map>  
#include <cmath>  
using namespace std;  
const int maxn = 200005;  
const int inf = 1111111111;  
long long q[maxn],head,tail,cost[maxn];  
long long p[maxn];  

int main()  
{  
    //freopen("in.txt","r",stdin);  
    int ca,n,m,use[maxn];  
    int u,v;  
    long long w;  
    bool flag;  
    scanf("%d",&ca);  
    while(ca--)  
    {  
        scanf("%d%d",&n,&m);  
        flag = 1;  
        for(int i = 1; i <= n; i++)  
        {  
            scanf("%d%d%d",&u,&v,&p[i]);  
            w=(long long)u*v;  
            if(w>m) flag = 0;  
            else cost[i]=u*v;  
        }  
        if(flag == 0)  
        {  
            printf("Impossible\n");  
            continue;  
        }  
        head = tail = 0;  
        long long ans = 0;  
        long long gas=0;  
        for(int i = 1; i <= n; i++)  
        {  
            while(head<tail&&p[i]<p[q[tail-1]]) gas-=use[--tail];  
            use[tail]=m - gas;  
            q[tail++]=i;  
            //cout<<gas<<" "<<m-gas<<endl;  
            gas = m - cost[i];  
            //cout<<gas<<endl;  
            while(cost[i])  
            {  
                if(use[head]<=cost[i])  
                {  
                    cost[i]-=use[head];  
                    ans+=use[head]*p[q[head]];  
                    head++;  
                }  
                else  
                {  
                    use[head]-=cost[i];  
                    ans+=cost[i]*p[q[head]];  
                    cost[i]=0;  
                }  
            }  
        }  
        cout<<ans<<endl;  
    }  
    return 0;  
}