题解 HDU 3698 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/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;

}