【原题题面】传送门
【题解大意】
当两个决策k1<k2且 f[i-1,k1] - p*k1 <= f[i-1,k2]-p*k2,那么此时k1就是无用决策。
可以用单调队列优化。
需要支持的操作:
1.当j变大时,b把小于j-L的决策出队;
2.有新的决策入队时,在队尾检查f[i-1,k]的单调性,把无用决策从队尾直接出队,然后把新决策入队。
【code】
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define File "fence" inline void file(){ freopen(File".in","r",stdin); freopen(File".out","w",stdout); } inline int read(){ int x = 0,f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){if(ch == '-')f = -1; ch = getchar();} while(ch >= '0' && ch <= '9'){x = (x<<1) + (x<<3) + ch-'0'; ch = getchar();} return x*f; } const int mxn = 16000 + 10; const int mxk = 105; int n,k; struct P{ int l,v,s; }p[mxk]; inline bool cmp(P t1,P t2){ return t1.s < t2.s; } int q[mxn]; int f[mxk][mxn]; inline int Calc(int i,int j){ return f[i-1][j] - p[i].v*j; } int main(){ file(); n = read(),k = read(); for(int i = 1;i <= k; ++i) p[i] = (P){read(),read(),read()}; sort(p+1,p+k+1,cmp); int l(1),r(0); for(int i = 1;i <= k; ++i){ l = 1,r = 0; for(int j = max(p[i].s-p[i].l,0);j <= p[i].s-1; ++j){ while(l <= r && Calc(i,q[r]) <= Calc(i,j)) --r; q[++r] = j;//新的决策从队尾入队 } for(int j = 1;j <= n; ++j){ f[i][j] = max(f[i-1][j],f[i][j-1]); if(j >= p[i].s){ while(l <= r && q[l] < j - p[i].l) ++l; if(l <= r) f[i][j] = max(f[i][j],Calc(i,q[l]) + p[i].v*j); } } } printf("%d\n",f[k][n]); return 0; }