zoj 2860 Breaking Strings
A certain string-processing language allows the programmer to break a string into two pieces. Since this involves copying the old string, it costs n units of time to break a string of n characters into two pieces. Suppose a programmer wants to break a string into many pieces. The order in which the breaks are made can affect the total amount of time used. For example, suppose we wish to break a 20 character string after characters 3, 8, and 10 (numbering the characters in ascending order from the left-hand end, starting from 1). If the breaks are made in left-to-right order, then the first break cost 20 units of time, the second break costs 17 units of time, and the third breaks costs 12 units of time, a total of 49 units of time (see the sample below). If the breaks are made in right-to-left order, then the first break costs 20 units of time, the second break costs 10 units of time, and the third break costs 8 units of time, a total of 38 units of time.
The cost of making the breaks in left-to-right order:
thisisastringofchars (original)
thi sisastringofchars (cost:20 units)
thi sisas tringofchars (cost:17 units)
thi sisas tr ingofchars (cost:12 units)
Total: 49 units.
The cost of making the breaks in right-to-left order:
thisisastringofchars (original)
thisisastr ingofchars (cost:20 units)
thisisas tr ingofchars (cost:10 units)
thi sisas tr ingofchars (cost: 8 units)
Total: 38 units.
Input:
There are several test cases! In each test case, the first line contains 2 integers N (2<=N<=10000000) and M (1<=M<=1000, M<N). N is the original length of the string, and M is the number of the breaks. The following lines contain M integers Mi (1<=Mi<N) in ascending order that represent the breaking positions from the string's left-hand end.
Output:
For each test case, output in one line the least cost to make all the breakings.
Sample Input:
20 3 3 8 10
Sample Output:
37Author: Wei, Qizheng
Source: ZOJ Monthly, June 2007
开始看到题目给的n发现数据量巨大如果用dp的话应该开不下。纠结一会儿后茅塞顿开。可以根据分割点来dp.若有m个分割点。那么整个字符串就被分割成0-1-2-3-.....-m-m+1个点而要分割的点在0-m+1之间。那么定义dp[i][j]为把i到j内的分割点分割好的最小花费。那么转移方程就显而易见了。
dp[i][j]=dp[i][k]+dp[k][j]+cost[i][j].注意是k不是k+1因为若是k+1若k+1是分割点dp[k+1][j]就没包含分割k+1点的价值了。
cost[i][j]=a[j]-a[i]即区间长度。
根据cost[i][j]的结构显然满足包含单调性。
设i'<i<j<j'。则有a[i']<a[i]<a[j]<a[j']。
设a[i']=a。a[i]=a+b。a[j]=a+b+c。a[j']=a+b+c+d。
cost[i'][j]=b+c。cost[i][j']=c+d。cost[i][j]=c。cost[i'][j']=b+c+d。
所以cost[i'][j]+cost[i][j']<=cost[i'][j']+cost[i][j].
所以满足四边形不等式。
所以k的枚举范围就变为s[i][j-1]<=k<=s[i+1][j]。缩小的枚举范围复杂度降到了o(n^2)。
#include <iostream>
#include<stdio.h>
#include<string.h>
#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define positive(a) ((a)>0?(a):-(a))
using namespace std;
int dp[1010][1010],cost[1010][1010],s[1010][1010],a[1010];
int n,m;
void solve()
{
int i,j,k,l,r,len,temp;
for(len=3;len<=m+1;len++)//枚举区间长度。保证可递推
{
for(i=0;i+len-1<=m;i++)//枚举i从大到小从小到大都可以
{
j=i+len-1;
l=s[i][j-1];//左边界
r=s[i+1][j];//右边界
for(k=l;k<=r;k++)//在区间内枚举k
{
temp=dp[i][k]+dp[k][j]+cost[i][j];
if(temp<dp[i][j])
{
dp[i][j]=temp;
s[i][j]=k;
}
}
}
}
}
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=m;i++)
scanf("%d",a+i);
a[0]=0;a[i]=n;
m++;//有m个分割点的话端点为m+1
memset(dp,0x3f,sizeof dp);
for(i=0;i<m;i++)//分割区间的花费为区间长度
for(j=i+1;j<=m;j++)
cost[i][j]=a[j]-a[i];
for(i=0;i<=m;i++)
{
s[i][i+1]=i+1;//区间长度为一的枚举范围为i到i+1
dp[i][i+1]=0;//区间长度为1时不用分割花费为0
cost[i][i+1]=0;
}
solve();
printf("%d\n",dp[0][m]);
}
return 0;
}