hdu_4283 You Are the One # by nobody
2012-09-12 20:19
387 查看
天津网赛一题简单的DP..
题意:给定n(n<=100)个人,每一个有固定的权值D,现在n个人从1~n排成一排,第i个人第j个出列的时候的unhappiness值是Di* (j-1),现在有一个栈,
可以暂时不让当前的人出列将其压栈,问怎样安排能使得sigma(unhappiness i)最小,求这个最小值。
解析:记忆化搜索dp[x][y]代表x-y人排队的最小sigma值,分第x个人第i次出栈得出最优子结构递归。
详见代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[110][110];
int s[110];
int d[110];
int dfs(int x,int y)
{
if(dp[x][y]!=-1)return dp[x][y];
if(x>=y)return dp[x][y] = 0;
int temp=1<<30;
for(int i=x;i<=y;i++)
{
temp = min(temp,dfs(x+1,i)+dfs(i+1,y)+(s[y]-s[i])*(i-x+1)+d[x]*(i-x));
}
return dp[x][y] = temp;
}
int main()
{
int cases;
scanf("%d",&cases);
int t = 1;
while(cases--)
{
int n;
memset(dp,-1,sizeof(dp));
scanf("%d",&n);
s[0] = 0;
for(int i=1;i<=n;i++)
{
scanf("%d",&d[i]);
s[i] = s[i-1]+d[i];
}
printf("Case #%d: %d\n",t++,dfs(1,n));
}
return 0;
}
题意:给定n(n<=100)个人,每一个有固定的权值D,现在n个人从1~n排成一排,第i个人第j个出列的时候的unhappiness值是Di* (j-1),现在有一个栈,
可以暂时不让当前的人出列将其压栈,问怎样安排能使得sigma(unhappiness i)最小,求这个最小值。
解析:记忆化搜索dp[x][y]代表x-y人排队的最小sigma值,分第x个人第i次出栈得出最优子结构递归。
详见代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[110][110];
int s[110];
int d[110];
int dfs(int x,int y)
{
if(dp[x][y]!=-1)return dp[x][y];
if(x>=y)return dp[x][y] = 0;
int temp=1<<30;
for(int i=x;i<=y;i++)
{
temp = min(temp,dfs(x+1,i)+dfs(i+1,y)+(s[y]-s[i])*(i-x+1)+d[x]*(i-x));
}
return dp[x][y] = temp;
}
int main()
{
int cases;
scanf("%d",&cases);
int t = 1;
while(cases--)
{
int n;
memset(dp,-1,sizeof(dp));
scanf("%d",&n);
s[0] = 0;
for(int i=1;i<=n;i++)
{
scanf("%d",&d[i]);
s[i] = s[i-1]+d[i];
}
printf("Case #%d: %d\n",t++,dfs(1,n));
}
return 0;
}
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