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): 821 Accepted Submission(s): 285
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
题意：就是每行选一个，上下两行需满足|j-k|≤f(i,j)+f(i+1,k).，问最小的cell和值。
分析：明显的dp,dp[i][j]表示到第i行选第j个的值，可是这样转移复杂度须要n*m*m，肯定会超时。
我们注意到|j-k|<=f(i,j)+f(i-1,k)，那么对于i-1行的第k个我们更新[k-f(i-1,k),k+f(i-1,k)]，对于第i行查询[j-f(i,j),j+f(i,j)]，这样刚好满足的是要求的条件。
所以就用线段树维护一下查询区间最小值更新区间值就好。这样复杂度就是n*m*log(m)。。
/**
* @author neko01
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-8;
const double pi=acos(-1.0);
const int INF=0x7fffffff;
const LL inf=(((LL)1)<<61)+5;
const int N=105;
const int M=5005;
int a[N][M];
int f[N][M];
int dp[N][M];
struct node{
int l,r;
int Min;
int col;
}tree[M*4];
void build(int x,int l,int r)
{
tree[x].l=l,tree[x].r=r;
tree[x].Min=INF;
tree[x].col=INF;
if(l==r) return;
int mid=(l+r)>>1;
build(x<<1,l,mid);
build(x<<1|1,mid+1,r);
}
inline void push_down(int x)
{
if(tree[x].col!=INF)
{
tree[x<<1].col=min(tree[x].col,tree[x<<1].col);
tree[x<<1|1].col=min(tree[x<<1|1].col,tree[x].col);
tree[x<<1].Min=min(tree[x].col,tree[x<<1].Min);
tree[x<<1|1].Min=min(tree[x].col,tree[x<<1|1].Min);
tree[x].col=INF;
}
}
void update(int x,int l,int r,int val)
{
if(tree[x].l==l&&tree[x].r==r)
{
tree[x].Min=min(tree[x].Min,val);
tree[x].col=min(tree[x].col,val);
return;
}
push_down(x);
int mid=(tree[x].l+tree[x].r)>>1;
if(r<=mid) update(x<<1,l,r,val);
else if(l>mid) update(x<<1|1,l,r,val);
else
{
update(x<<1,l,mid,val);
update(x<<1|1,mid+1,r,val);
}
tree[x].Min=min(tree[x<<1].Min,tree[x<<1|1].Min);
}
int query(int x,int l,int r)
{
if(tree[x].l==l&&tree[x].r==r)
return tree[x].Min;
push_down(x);
int mid=(tree[x].l+tree[x].r)>>1;
if(r<=mid) return query(x<<1,l,r);
else if(l>mid) return query(x<<1|1,l,r);
else return min(query(x<<1,l,mid),query(x<<1|1,mid+1,r));
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0) break;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%d",&a[i][j]);
if(i==1) dp[1][j]=a[i][j];
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&f[i][j]);
for(int i=2;i<=n;i++)
{
build(1,1,m);
for(int j=1;j<=m;j++)
{
int l=max(1,j-f[i-1][j]);
int r=min(m,j+f[i-1][j]);
update(1,l,r,dp[i-1][j]);
}
for(int j=1;j<=m;j++)
{
int l=max(1,j-f[i][j]);
int r=min(m,j+f[i][j]);
dp[i][j]=query(1,l,r)+a[i][j];
}
}
int ans=INF;
for(int i=1;i<=m;i++)
ans=min(ans,dp[n][i]);
printf("%d\n",ans);
}
return 0;
}