http://acm.hdu.edu.cn/showproblem.php?pid=3698

Let the light guide us

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 759    Accepted Submission(s): 253

Problem Description

Plain of despair was once an ancient battlefield where those brave spirits had rested in peace for thousands of years. Actually no one dare step into this sacred land until the rumor that “there is a huge gold mine underneath the plain” started to spread.
Recently an accident destroyed the eternal tranquility. Some greedy fools tried using powerful bombs to find the hidden treasure. Of course they failed and such behavior enraged those spirits--the consequence is that all the human villages nearby are haunted by ghosts.
In order to stop those ghosts as soon as possible, Panda the Archmage and Facer the great architect figure out a nice plan. Since the plain can be represented as grids of N rows and M columns, the plan is that we choose ONLY ONE cell in EACH ROW to build a magic tower so that each tower can use holy light to protect the entire ROW, and finally the whole plain can be covered and all spirits can rest in peace again. It will cost different time to build up a magic tower in different cells. The target is to minimize the total time of building all N towers, one in each row.
“Ah, we might have some difficulties.” said Panda, “In order to control the towers correctly, we must guarantee that every two towers in two consecutive rows share a common magic area.”
“What?”
“Specifically, if we build a tower in cell (i,j) and another tower in cell (i+1,k), then we shall have |j-k|≤f(i,j)+f(i+1,k). Here, f(i,j) means the scale of magic flow in cell (i,j).”
“How?”
“Ur, I forgot that you cannot sense the magic power. Here is a map which shows the scale of magic flows in each cell. And remember that the constraint holds for every two consecutive rows.”
“Understood.”
“Excellent! Let’s get started!”
Would you mind helping them?

Input

There are multiple test cases.
Each test case starts with a line containing 2 integers N and M (2<=N<=100,1<=M<=5000), representing that the plain consists N rows and M columns.
The following N lines contain M integers each, forming a matrix T of N×M. The j-th element in row i (Tij) represents the time cost of building a magic tower in cell (i, j). (0<=Tij<=100000)
The following N lines contain M integers each, forming a matrix F of N×M. The j-th element in row i (Fij) represents the scale of magic flows in cell (i, j). (0<=Fij<=100000)
For each test case, there is always a solution satisfying the constraints.
The input ends with a test case of N=0 and M=0.

Output

For each test case, output a line with a single integer, which is the minimum time cost to finish all magic towers.

Sample Input

3 5

9 5 3 8 7

8 2 6 8 9

1 9 7 8 6

0 1 0 1 2

1 0 2 1 1

0 2 1 0 2

0 0

Sample Output

10

Source

2010 Asia Fuzhou Regional Contest

分析:

这题显然是Dp .....dp[i][j] = min{dp[i + 1][k] + T[i][j]} |j-k|≤f(i,j)+f(i+1,k)

可是,一看数据规模,time有点捉急额、、、

斜率优化? .....没看出来。

四边形不等式? .....还是算了吧。

看来要优化状态转移方程“有点”困难......

让我们换个思路,能不能快速取到合适的k…

|j-k|≤f(i,j)+f(i+1,k)

Dp[i][j]的最优值,一定在以j为圆心,f[i][j]为半径的区间内

而dp[i-1][j]能更新以j为圆心,f[i-1][j] 为半径的区间。

这样,我们用每个dp[i-1][j] 更新每个[j – f[i - 1][j] , j + f[i - 1][j]]范围内的值。

计算 Dp[i][j]时,只需查询[j – f[i][j] , j + f[i][j]]范围内的最小值,既RMQ

So。。。。看代码吧。。。。

AC Code:    呵呵,在HDU上排名第13.......

#include <cstdio>

using namespace std;

const int maxn = 105;

const int maxm = 5005;

const int inf = 1 << 30;

inline int L(int rt)    {return rt << 1;}

inline int R(int rt)    {return rt << 1 | 1;}

inline int min(int a,int b) {return a < b ? a : b;}

struct Node

{

    int l,r;

    int min;        // 区间最小值

    int val;

} nd[maxm << 2];

int f[maxn][maxm];

int time[maxn][maxm];

int dp[maxn][maxm];

inline void pushDown(int rt)

{

    if(nd[rt].val != inf)                            // val 有修改

    {

        nd[L(rt)].val = min(nd[L(rt)].val,nd[rt].val);

        nd[R(rt)].val = min(nd[R(rt)].val,nd[rt].val);

        nd[L(rt)].min = min(nd[L(rt)].min,nd[L(rt)].val);

        nd[R(rt)].min = min(nd[R(rt)].min,nd[R(rt)].val);

        nd[rt].val = inf;

    }

}

void build(int l,int r,int rt)                            // 建线段树

{

    nd[rt].l = l;   nd[rt].r = r;

    nd[rt].min = nd[rt].val = inf;

    if(l == r)  return;

    int mid = (l + r) >> 1;

    build(l,mid,L(rt));

    build(mid + 1,r,R(rt));

}

void update(int l,int r,int val,int rt)                     // 更新区间[l,r]

{

    if(l <= nd[rt].l && nd[rt].r <= r)

    {

        nd[rt].val = min(val,nd[rt].val);

        nd[rt].min = min(nd[rt].val,nd[rt].min);

        return;

    }

    int mid = (nd[rt].l + nd[rt].r) >> 1;

    pushDown(rt);

    if(l <= mid)    update(l,r,val,L(rt));

    if(r > mid)     update(l,r,val,R(rt));

    nd[rt].min = min(nd[L(rt)].min,nd[R(rt)].min);

}

int query(int l,int r,int rt)                            // RMQ

{

    if(l <= nd[rt].l && nd[rt].r <= r)

        return nd[rt].min;

    int ret = inf;

    int mid = (nd[rt].l + nd[rt].r) >> 1;

    pushDown(rt);

    if(l <= mid)    ret = min(ret,query(l,r,L(rt)));

    if(r > mid)     ret = min(ret,query(l,r,R(rt)));

    return ret;

}

inline void scan(int &n)                     // 读入加速

{

    char c;

    while(c = getchar(),c < '0' || c > '9');

    n = c - '0';

    while(c = getchar(),(c >= '0' && c <= '9')) n = n * 10 + c - '0';

}

int main()

{

    int n,m;

    while(scanf("%d%d",&n,&m) && (n != 0 || m != 0))

    {

        for(int i = 1;i <= n;i ++)

        for(int j = 1;j <= m;j ++)

            scan(time[i][j]);

        for(int i = 1;i <= n;i ++)

        for(int j = 1;j <= m;j ++)

            scan(f[i][j]);

        for(int i = 1;i <= m;i ++)              // 初始化边界

            dp[1][i] = time[1][i];

/////////////////////////////////////////////////////核心/////////////////////////////////////////////////////

        for(int i = 2;i <= n;i ++)

        {

            build(1,m,1);

            for(int j = 1;j <= m;j ++)

                update(j - f[i - 1][j],j + f[i - 1][j],dp[i - 1][j],1);

            for(int j = 1;j <= m;j ++)

                dp[i][j] = query(j - f[i][j],j + f[i][j],1) + time[i][j];

        }

/////////////////////////////////////////////////////END/////////////////////////////////////////////////////

        int ans = inf;

        for(int i = 1;i <= m;i ++)

            ans = min(ans,dp[n][i]);

        printf("%d\n",ans);

    }

    return 0;

}

题解 HDU 3698 Let the light guide us Dp + 线段树优化的更多相关文章

  1. hdu3698 Let the light guide us dp+线段树优化

    http://acm.hdu.edu.cn/showproblem.php?pid=3698 Let the light guide us Time Limit: 5000/2000 MS (Java ...

  2. HDU 3698 Let the light guide us

    Let the light guide us Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on HDU. ...

  3. hdu 3698 Let the light guide us(线段树优化&简单DP)

    Let the light guide us Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/O ...

  4. HDU 3698 Let the light guide us(DP+线段树)(2010 Asia Fuzhou Regional Contest)

    Description Plain of despair was once an ancient battlefield where those brave spirits had rested in ...

  5. HDU 4719 Oh My Holy FFF(DP+线段树)(2013 ACM/ICPC Asia Regional Online ―― Warmup2)

    Description N soldiers from the famous "*FFF* army" is standing in a line, from left to ri ...

  6. hdu 3698 UVA1490 Let the light guide us 线段树优化DP

    题目链接 and 题目大意 hdu3698 但是 hdu的数据比较弱,所以在这luogu提交吧UVA1490 Let the light guide us 有一个\(n*m\)的平原,要求每行选一个点 ...

  7. hdu 5266 pog loves szh III(lca + 线段树)

    I - pog loves szh III Time Limit:6000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I ...

  8. hdu 3016 dp+线段树

    Man Down Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total S ...

  9. HDU 2795 Billboard(宣传栏贴公告,线段树应用)

    HDU 2795 Billboard(宣传栏贴公告,线段树应用) ACM 题目地址:HDU 2795 Billboard 题意:  要在h*w宣传栏上贴公告,每条公告的高度都是为1的,并且每条公告都要 ...

随机推荐

  1. MacOS最佳思维导图推荐-MindNode 7

    思维导图软件哪个比较好呢?MindNode for mac下载一款功能简单,界面简洁,不用看教程都会用的思维导图软件.mindnode 7 mac版可随时随地记录自己的想法,让您从灵感入手,将奇思妙想 ...

  2. [原]排错实战——解救加载调试符号失败的IDA

    原调试IDA排错troubleshootsymbolspdbsysinternalprocess monitor 缘起 最近想借助IDA逆向一个函数.在windows下,调试器(比如vs, windb ...

  3. 将iso mount 到nfs 目录问题

    最近有个需求,需要在多台系统安装程序,安装文件是iso 格式的,最普通的办法就是拷贝到其它系统,然后mount loop 到本地目录. 但是比较麻烦,而且当前已经有一个nfs 服务端了,于是想出一个办 ...

  4. 题解【语文1(chin1)- 理理思维】

    link 喵~珂朵莉树AC 珂朵莉树?见此处~ 这数据结构太暴力了,所以不讲了 Code: #include<iostream> #include<cstdio> #inclu ...

  5. 1013A.Piles With Stones

    题目出处:http://codeforces.com/contest/1013/problem/A #include<iostream> using namespace std; int ...

  6. requset请求处理与BeanUtils封装

    HTTP: 概念:Hyper Text Transfer Protocol 超文本传输协议 传输协议:定义了,客户端和服务器端通信时,发送数据的格式 特点: 基于TCP/IP的高级协议 默认端口号:8 ...

  7. 蓝桥杯 传球游戏(dp)

    Description 上体育课的时候,小蛮的老师经常带着同学们一起做游戏.这次,老师带着同学们一起做传球游戏.游戏规则是这样的:n个同学站成一个圆圈,其中的一个同学手里拿着一个球,当老师吹哨子时开始 ...

  8. Flink(二) —— 部署与任务提交

    一.下载&启动 官网上下载安装包,执行下列命令即启动完成. ./bin/start-cluster.sh 效果图 Flink部署模式 Standalone模式 Yarn模式 k8s部署 二.配 ...

  9. vue项目中的elementUI的table组件导出成excel表

    1.安装依赖:npm install --save xlsx file-saver 2.在放置需要导出功能的组件中引入 import FileSaver from 'file-saver' impor ...

  10. linux下特殊命令集锦

    1.ifconfig -a | grep enp0 | cut -d : -f 1   //按照:进行切割网络文件名 如:ifconfig `ifconfig -a | grep enp0 | cut ...