HDU

题目：将一个序列分成m+1段，每段的价值定义为任意两个数的乘积之和，求总的最小价值

思路：

cost[i][j]表示i到j成为一段的价值

设dp[i][j]表示前j个数分成i段的最小价值

dp[i][j]=min{dp[i-1][k]+cost[k+1][j]}

cost[1][i]=cost[1][k]+cost[k+1][i]+sum[k]*(sum[i]-sum[k])

将cost[k+1][i]代入上面的式子，得

dp[i][j]=min{dp[i-1][k]+cost[1][j]-cost[1][k]-sum[k]*(sum[j]-sum[k])}

然后斜率优化即可

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }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;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int maxn=1e3+5;int arr[maxn],sum[maxn],cost[maxn];int dp[maxn][maxn];int q[maxn];int head,tail,n,m;int getdp(int i,int j,int k){    return dp[i-1][k]+cost[j]-cost[k]-sum[k]*(sum[j]-sum[k]);}int gety(int i,int k,int j){//yj-yk    return dp[i-1][j]-cost[j]+sum[j]*sum[j]-(dp[i-1][k]-cost[k]+sum[k]*sum[k]);}int getx(int k,int j){//xj-xk    return sum[j]-sum[k];}int main(){//    freopen("D:\\input.txt","r",stdin);//    freopen("D:\\output.txt","w",stdout);    while(~scanf("%d%d",&n,&m)){        if(n==0&&m==0)            break;        sum[0]=0;        for(int i=1;i<=n;i++){            scanf("%d",&arr[i]);            sum[i]=sum[i-1]+arr[i];            cost[i]=cost[i-1]+sum[i-1]*arr[i];        }        for(int i=1;i<=n;i++)            dp[1][i]=cost[i];        m++;        for(int i=2;i<=m;i++){            head=0;            tail=-1;            q[++tail]=i-1;            for(int j=i;j<=n;j++){                while(head<tail&&gety(i,q[head],q[head+1])<=sum[j]*getx(q[head],q[head+1]))                    head++;                dp[i][j]=getdp(i,j,q[head]);                while(head<tail&&gety(i,q[tail-1],q[tail])*getx(q[tail],j)>=gety(i,q[tail],j)*getx(q[tail-1],q[tail]))                    tail--;                q[++tail]=j;            }        }        printf("%d\n",dp[m][n]);    }    return 0;}