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？。。。
1 //STATUS:C++_AC_1172MS_1352KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 typedef __int64 LL; 33 typedef unsigned __int64 ULL; 34 //const 35 const int N=100010; 36 const int INF=0x3f3f3f3f; 37 const int MOD=100000,STA=8000010; 38 const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 int num[N],q[N]; 58 int n; 59 LL m,f[N]; 60 multiset<int> sbt; 61 62 int main() 63 { 64 // freopen("in.txt","r",stdin); 65 int i,j,l,r,p,ok; 66 LL sum; 67 while(~scanf("%d%I64d",&n,&m)) 68 { 69 l=sum=0;r=-1; 70 sbt.clear(); 71 ok=1; 72 for(i=p=1;i<=n;i++){ 73 scanf("%d",&num[i]); 74 sum+=num[i]; 75 while(sum>m)sum-=num[p++]; 76 if(p>i){ok=0;break;} 77 while(l<=r && num[i]>=num[q[r]]){ 78 if(l<r)sbt.erase(f[q[r-1]]+num[q[r]]); 79 r--; 80 } 81 q[++r]=i; 82 if(l<r)sbt.insert(f[q[r-1]]+num[q[r]]); 83 while(q[l]<p){ 84 if(l<r)sbt.erase(f[q[l]]+num[q[l+1]]); 85 l++; 86 } 87 f[i]=f[p-1]+num[q[l]]; 88 if(l<r)f[i]=Min(f[i],(LL)*sbt.begin()); 89 } 90 for(;i<=n;i++) 91 scanf("%d",&j); 92 93 printf("%I64d\n",ok?f[n]:-1); 94 } 95 return 0; 96 }