- 区间DP模板:三层循环
循环1 --------- 枚举区间长度
循环2 --------- --------- 枚举区间开始位置(同时计算出区间结束位置)
循环3 --------- --------- --------- 枚举区间分割点k,并结合状态转移方程进行DP - 本题还可用四边形优化,即:s(i,j-1)<=s(i,j)<=s(i+1,j) (s(i,j)=k)
未用四边形优化代码
#include<iostream>
#include<cstdio>
#include<climits>
using namespace std;
const int N = 1001;
int sum[N],t[N];
int dp[N][N],p[N][N];
int main()
{
int n;
scanf("%d", &n);
sum[0] = 0;
fill(dp[0], dp[0]+N*N, INT_MAX);
for(int i = 1; i <= n; i++){
scanf("%d", &t[i]);
sum[i] = sum[i-1] + t[i];
dp[i][i] = 0;
//p[i][i] = i;
}
for(int len = 2; len <= n; len++){
for(int i = 1; i <= n; i++){
int j = i+len-1;
if(j > n) continue;
//for(int k = p[i][j-1]; k <= p[i+1][j]; k++){
for(int k = i; k < j; k++){
int tmp = dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1];
if(dp[i][j] > tmp){
dp[i][j] = tmp;
//p[i][j] = k;
}
}
}
}
printf("%d", dp[1][n]);
return 0;
}
用了四边形优化后代码
#include<iostream>
#include<cstdio>
#include<climits>
#include<cstring>
using namespace std;
const int N = 1001;
const int INT_MAX2 = 0x7F7F7F7F;
int sum[N],t[N];
int dp[N][N],p[N][N];
int main()
{
int n;
scanf("%d", &n);
sum[0] = 0;
memset(dp, INT_MAX2, sizeof(dp));
//fill(dp[0], dp[0]+N*N, INT_MAX);
for(int i = 1; i <= n; i++){
scanf("%d", &t[i]);
sum[i] = sum[i-1] + t[i];
dp[i][i] = 0;
p[i][i] = i;
}
for(int len = 2; len <= n; len++){
for(int i = 1; i <= n; i++){
int j = i+len-1;
if(j > n) continue;
for(int k = p[i][j-1]; k <= p[i+1][j]; k++){
int tmp = dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1];
if(dp[i][j] > tmp){
dp[i][j] = tmp;
p[i][j] = k;
}
}
}
}
printf("%d", dp[1][n]);
return 0;
}