HDU4283You Are the One区间dp，
记忆话搜索运行时间：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 105,inf = 1<<30;
int dp[maxn][maxn][maxn],arr[maxn],n;
int solve(int s,int e,int k){
if( s == e)return arr[s] * (k-1);
if( s > e)return 0;
if(dp[s][e][k] != inf)return dp[s][e][k]; for(int i = s; i <= e; i++){ //将区间[s,e]划分为[s,i],[i+1,e]
int nextk = k+(i-s)+1; //[i+1,e]区间内第i个人第nextk个出栈
int thisk = k+(i-s); //[s,i]区间的第s个人第thisk个出栈
int first = solve(s+1,i,k); //第1个区间经过分配得到最少屌丝值
int second = solve(i+1,e,nextk); //第2个区间经过分配得到的最少屌丝值
int cur = arr[s] * (thisk-1); //第s个人第thisk出栈增加的屌丝值
dp[s][e][k] = min(dp[s][e][k],cur+second+first);
}
return dp[s][e][k];
}
int main(){
int t,cas = 1;
scanf("%d",&t);
while( t-- ){
scanf("%d",&n);
for(int i = 1; i <= n; i++)
scanf("%d",arr+i);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
for(int k = 0; k <= n; k++)
dp[i][j][k] = inf;
printf("Case #%d: %d\n",cas++,solve(1,n,1)); }
return 0;
}
三个for循环
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn =105;
const int inf = 1<<30;
int dp[maxn][maxn],arr[maxn],sum[maxn];
int n;
int solve_dp(){
for(int len = 1; len < n; len++){ //枚举区间长度
for(int s = 1; s + len <= n; s ++){ //枚举区间起点
int end = s+len; //s为区间头，end为区间尾
for(int k = s; k <= end; k++){ //第k个出栈
int tp = arr[s] * (k-s);
tp += dp[s+1][k] + dp[k+1][end];
tp += (k-s+1) * (sum[end] - sum[k]);//重要的性质，如果第i个人第k个出栈，那么后面的人出栈顺序都大于k
dp[s][end] = min(dp[s][end],tp);
}
}
}
return dp[1][n];
}
int main(){
int t,cas = 1;
scanf("%d",&t);
while( t-- ){
scanf("%d",&n);
for(int i = 1; i <= n; i++){
scanf("%d",arr+i);
sum[i] = sum[i-1] + arr[i];
}
memset(dp,0,sizeof(dp));
for(int i = 1; i <= n; i++)
for(int j = i+1; j <= n; j++)
dp[i][j] = inf;
printf("Case #%d: %d\n",cas++,solve_dp());
}
return 0;
}
经典的区间DP模型--最大括号匹配数。如果找到一对匹配的括号[xxx]oooo，就把区间分成两部分，一部分是xxx，一部分是ooo，然后以此递归直到区间长度为1或者为2.
状态转移方程:if(mach(j,j+i))dp[j][j+i] = dp[j+1][j+i-1]+2;
dp[j][j+i] = max(dp[j][k],dp[k+1][j+i])(j <= k < j+i)
/*
** 一种自底向上型的DP
** 状态转移方程：if(mach(j,j+i))dp[j][j+i] = dp[j+1][j+i-1]+2;
** dp[j][j+i] = max(dp[j][k],dp[k+1][j+i])(j <= k < j+i)
*/
#include <stdio.h>
#include <cstring>
#define Max(a,b) (a)>(b)?(a):(b)
#define maxn 101
int dp[maxn][maxn];
char str[maxn];
int main()
{
while(~scanf("%s",str)) {
if(strcmp(str,"end") == 0)break;
int len = strlen(str);
memset(dp,0,sizeof(dp));
for(int i = 1; i <= len; i++) {//枚举区间长度
for(int j = 0; j <= len - i; j++) {//枚举起点
dp[j][j+i] = dp[j+1][j+i-1] ;
if((str[j] == '(' && ')'== str[j+i]) || (str[j] =='[' && str[j+i] == ']' ))
dp[j][j+i] += 2;
for(int k = j; k < i+j; k++)//枚举区间j->j+i的分界点
dp[j][j+i] = Max(dp[j][j+i],dp[j][k]+dp[k+1][j+i]);
}
}
printf("%d\n",dp[0][len-1]);
}
return 0;
}
Light OJ 1422 Halloween Costumes
很基础的区间DP,是不老传说那题的减弱版。
状态转移方程：
dp[s][e] = dp[s][e-1] (arr[s] == arr[e]).dp[s][e] = dp[s][e-1]+1 (arr[s] != arr[e]) ;
dp[s][e] = min(dp[s][e] ,dp[s][k]+dp[k+1][e]);(arr[s] == arr[k] && s <= k < e )
dp[s][e] = min(dp[s][e] ,dp[s][k]+dp[k+1][e]+1);(arr[s] != arr[k] && s <= k < e)
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 110;
int dp[maxn][maxn],arr[maxn];
int main()
{
int t,cas = 1,n;
scanf("%d",&t);
while( t-- ) {
scanf("%d",&n);
for(int i = 1; i <= n; i++)
scanf("%d",&arr[i]);
memset(dp,0,sizeof(dp));
for(int i = 1; i <= n; i++)dp[i][i] = 1;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n-i; j++) {
dp[j][j+i] = dp[j][j+i-1];
if(arr[j] != arr[j+i])dp[j][j+i]++;//当起点和终点不同时衣服数量加1
for(int k = j; k < j+i; k++) {
if(arr[j] == arr[k])
dp[j][j+i] = min(dp[j][j+i],dp[j][k]+dp[k+1][j+i]);//当第j和第k相等时，衣服数量等于(j->k) + (k+1->j+i)
else
dp[j][j+i] = min(dp[j][j+i],dp[j][k]+dp[k+1][j+i]+1);//当j和k不相等时，衣服数量等于(j->k) + (k+1->j+i) + 1
}
}
printf("Case %d: %d\n",cas++,dp[1][n]);
}
return 0;
}