优先队列贪心

最近做了几个用优先队列进行贪心的题目，这里写下来，以免忘了。
1、CF 799 B http://codeforces.com/problemset/problem/799/B
每次选最便宜的衣服，没什么好说的，一发过
代码：

#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 1000000007double 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‘？’都换成右括号，放进队列，当不满足对称时选cost最小的就可以了
代码：

#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 1000000007double 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
经典的加油问题
代码：

#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;  }