本文共 2283 字，大约阅读时间需要 7 分钟。
题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1010
题意：给出一个数列，一个数字L，将数列分成若干段，s[i]为前i项和，某段[i,j]的代价为(s[i]-s[j-1]+i-j-L)^2。最后使得总代价最小？
思路：单调队列维护斜率。
#include #include #include #include #include #include #include #include #include #include #include #define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)<(y)?(x):(y))#define abs(x) ((x)>=0?(x):-(x))#define i64 long long#define u32 unsigned int#define u64 unsigned long long#define clr(x,y) memset(x,y,sizeof(x))#define CLR(x) x.clear()#define ph(x) push(x)#define pb(x) push_back(x)#define Len(x) x.length()#define SZ(x) x.size()#define PI acos(-1.0)#define sqr(x) ((x)*(x))#define MP(x,y) make_pair(x,y)#define EPS 1e-9#define FOR0(i,x) for(i=0;i =0;i--)#define FORL1(i,a) for(i=a;i>=1;i--)#define FORL(i,a,b)for(i=a;i>=b;i--)#define rush() int CC;for(scanf("%d",&CC);CC--;)#define Rush(n) while(scanf("%d",&n)!=-1)using namespace std;void RD(int &x){scanf("%d",&x);}void RD(u32 &x){scanf("%u",&x);}void RD(i64 &x){scanf("%lld",&x);}void RD(double &x){scanf("%lf",&x);}void RD(int &x,int &y){scanf("%d%d",&x,&y);}void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}void RD(i64 &x,i64 &y,i64 &z){scanf("%lld%lld%lld",&x,&y,&z);}void RD(int &x,int &y,int &z,int &t){scanf("%d%d%d%d",&x,&y,&z,&t);}void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}void RD(char &x){x=getchar();}void RD(char *s){scanf("%s",s);}void RD(string &s){cin>>s;}void PR(int x) {printf("%d\n",x);}void PR(int x,int y) {printf("%d %d\n",x,y);}void PR(int x,int y,int z) {printf("%d %d %d\n",x,y,z);}void PR(i64 x) {printf("%lld\n",x);}void PR(u32 x) {printf("%u\n",x);}void PR(double x) {printf("%.5lf\n",x);}void PR(char x) {printf("%c\n",x);}void PR(char *x) {printf("%s\n",x);}void PR(string x) {cout< < =(s[i]+i)*dx(Q[head],Q[head+1])) head++; j=Q[head]; dp[i]=dp[j]+sqr(s[i]-s[j]+i-j-1-L); z=i; while(head =dy(y,z)*dx(x,y)) tail--; else break; } Q[++tail]=z; } PR(dp[n]);}int main(){ Rush(n) { RD(L); int i; FOR1(i,n) RD(s[i]),s[i]+=s[i-1]; DP(); } return 0;}
转载地址：http://syxal.baihongyu.com/