POJ-3017 Cut the Sequence DP+单调队列+堆

　　题目链接：http://poj.org/problem?id=3017

　　这题的DP方程是容易想到的，f[i]=Min{ f[j]+Max(num[j+1],num[j+2],......,num[i]) | 满足m的下界<j<=i }，复杂度O(n^2)，妥妥的TLE。其实很多都决策都是没有必要的，只要保存在满足m的区间内，num值单调递减的的那些决策。如果遍历的话，一个下降的序列会退化到O(n^2)，于是用堆来优化。。。堆优化这里，纠结了很久T_T，，，网上很多代码都是直接用set来处理，但是set在erase元素的都是会把相同的元素都除掉，应该是只erase一个元素，因为相同的元素中其它的可能会存在队列中。。。难道是数据弱了？。。。

 //STATUS:C++_AC_1172MS_1352KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int num[N],q[N];
 int n;
 LL m,f[N];
 multiset<int> sbt;

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,l,r,p,ok;
     LL sum;
     while(~scanf("%d%I64d",&n,&m))
     {
         l=sum=;r=-;
         sbt.clear();
         ok=;
         for(i=p=;i<=n;i++){
             scanf("%d",&num[i]);
             sum+=num[i];
             while(sum>m)sum-=num[p++];
             if(p>i){ok=;break;}
             while(l<=r && num[i]>=num[q[r]]){
                 if(l<r)sbt.erase(f[q[r-]]+num[q[r]]);
                 r--;
             }
             q[++r]=i;
             if(l<r)sbt.insert(f[q[r-]]+num[q[r]]);
             while(q[l]<p){
                 if(l<r)sbt.erase(f[q[l]]+num[q[l+]]);
                 l++;
             }
             f[i]=f[p-]+num[q[l]];
             if(l<r)f[i]=Min(f[i],(LL)*sbt.begin());
         }
         for(;i<=n;i++)
             scanf("%d",&j);

         printf("%I64d\n",ok?f[n]:-);
     }
     return ;
 }