POJ-3017 Cut the Sequence DP+单调队列+堆

　　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 ;

 }

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POJ-3017 Cut the Sequence DP+单调队列+堆

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