【DP】 HDU 5115 Dire Wolf 区间DP

点击打开链接

题意:一头狼的攻击力= 自己攻击力+相邻两边的狼的加成

每杀一头狼会收到一次攻击

求受到的攻击和最小

比较裸的区间DP

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <deque>
#include <map>
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long  LL;
const int MAXN = 230+1;
const int MAXM = 140000;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
int a[3000],b[3000];
int dp[300][300];
int dfs(int l,int r)
{
    if(l+1>=r) return 0;
    if(dp[l][r]!=-1) return dp[l][r];
    dp[l][r]=INF;
    for(int i=l+1;i<r;i++)
    {
        dp[l][r]=min(dp[l][r],dfs(l,i)+dfs(i,r)+a[i]+b[l]+b[r]);
    }
    return dp[l][r];
}
int main()
{
#ifndef ONLINE_JUDGE
       freopen("in.txt", "r", stdin);
   //  freopen("out.txt", "w", stdout);
#endif
    int t,n,cas=1;
    cin>>t;
    while(t--)
    {
        cin>>n;
        cler(dp,-1);
        a[0]=b[0]=a[n+1]=b[n+1]=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&b[i]);
        printf("Case #%d: %d\n",cas++,dfs(0,n+1));
    }
    return 0;
}
时间: 2024-10-23 13:15:15

【DP】 HDU 5115 Dire Wolf 区间DP的相关文章

HDU 5115 Dire Wolf 区间DP

Dire Wolf Problem Description Dire wolves, also known as Dark wolves, are extraordinarily large and powerful wolves. Many, if not all, Dire Wolves appear to originate from Draenor.Dire wolves look like normal wolves, but these creatures are of nearly

HDOJ 5115 Dire Wolf 区间DP

区间DP Dire Wolf Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 44    Accepted Submission(s): 30 Problem Description Dire wolves, also known as Dark wolves, are extraordinarily large and power

HDU 5115 Dire Wolf

上次北京赛现场赛的题了,昨天做了道区间dp,突然想起来这道题,都是很类似的,就翻出来做了做 刚开始像昨天做的那道一样,老想着怎么逆推,后来发现这道题应该正着推 其实正推和逆推乍看起来是很相似的,只不过一个是dp[i][j]表示i.j左右还有其他狼时消灭掉i-j这段消耗的费用,一个是表示最后只剩i-j时消灭掉这段的费用 总结一下昨天那道区间dp和今天的区间dp的相同点,发现区间dp适用于费用会随着时间的推移,或者说dp的进行不断变化 #include<stdio.h> #include<a

hdu 5115 Dire Wolf(区间dp)

Problem Description Dire wolves, also known as Dark wolves, are extraordinarily large and powerful wolves. Many, if not all, Dire Wolves appear to originate from Draenor. Dire wolves look like normal wolves, but these creatures are of nearly twice th

动态规划(区间DP):HDU 5115 Dire Wolf

Dire wolves, also known as Dark wolves, are extraordinarily large and powerful wolves. Many, if not all, Dire Wolves appear to originate from Draenor. Dire wolves look like normal wolves, but these creatures are of nearly twice the size. These powerf

HDU 5115 Dire Wolf ——(区间DP)

比赛的时候以为很难,其实就是一个区间DP= =..思路见:点我. 区间DP一定要记住先枚举区间长度啊= =~!因为区间dp都是由短的区间更新长的区间的,所以先把短的区间更新完.. 代码如下: 1 #include <stdio.h> 2 #include <algorithm> 3 #include <string.h> 4 using namespace std; 5 const int N = 200 + 5; 6 const int inf = 0x3f3f3f3

UVAlive 10154:Dire Wolf 区间DP

Dire Wolf 题目链接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5073 题意: 有一排的狼,每只狼有一个基础攻击力a[i],和一个加成值b[i](可以给相邻的两只狼加b[i]点攻击力,这只狼死后就不加了),你每杀一只狼需要花费能量值(狼的当前攻击力),你杀了这只狼周围的狼会自动往中间聚集(即不会有空隙),求杀

HDU5115 Dire Wolf (区间DP)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5115 题意: 有n只狼,每只狼有两种属性,一种攻击力一种附加值,我们没杀一只狼,那么我们受到的伤害值为 这只狼的攻击值与它旁边的两只狼的附加值的和,求把所有狼都杀光受到的最小的伤害值. 分析: 赤裸裸的区间DP dp[i][j]表示把区间i,j内的所有狼杀光所受到的最小的伤害. 状态转移方程为 dp[i][j]=min{dp[i][k]+dp[k+1][j]-b[k]+b[i+1],dp[i][k

Dire Wolf(区间DP)

Dire Wolf Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 2221    Accepted Submission(s): 1284 Problem Description Dire wolves, also known as Dark wolves, are extraordinarily large and powerfu