HDU 3698 Let the light guide us

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 105;
 4 const int M = 5005;
 5 const int INF = 0x3f3f3f3f;
 6 struct node {
 7     int minv,lazy;
 8 } tree[M<<2];
 9 int T[N][M],F[N][M],dp[N][M];
10 void build(int L,int R,int v) {
11     tree[v].lazy = INF;
12     tree[v].minv = INF;
13     if(L == R) return;
14     int mid = (L + R)>>1;
15     build(L,mid,v<<1);
16     build(mid+1,R,v<<1|1);
17 }
18 inline void pushdown(int v) {
19     if(tree[v].lazy < INF) {
20         tree[v<<1].lazy = min(tree[v<<1].lazy,tree[v].lazy);
21         tree[v<<1].minv = min(tree[v<<1].minv,tree[v<<1].lazy);
22         tree[v<<1|1].lazy = min(tree[v<<1|1].lazy,tree[v].lazy);
23         tree[v<<1|1].minv = min(tree[v<<1|1].minv,tree[v<<1|1].lazy);
24         tree[v].lazy = INF;
25     }
26 }
27 inline void pushup(int v) {
28     tree[v].minv = min(tree[v<<1].minv,tree[v<<1|1].minv);
29 }
30 void update(int L,int R,int lt,int rt,int val,int v) {
31     if(lt <= L && rt >= R) {
32         tree[v].lazy = min(tree[v].lazy,val);
33         tree[v].minv = min(tree[v].lazy,tree[v].minv);
34         return;
35     }
36     pushdown(v);
37     int mid = (L + R)>>1;
38     if(lt <= mid) update(L,mid,lt,rt,val,v<<1);
39     if(rt > mid) update(mid+1,R,lt,rt,val,v<<1|1);
40     pushup(v);
41 }
42 int query(int L,int R,int lt,int rt,int v) {
43     if(lt <= L && rt >= R) return tree[v].minv;
44     pushdown(v);
45     int mid = (L + R)>>1,ret = INF;
46     if(lt <= mid) ret = query(L,mid,lt,rt,v<<1);
47     if(rt > mid) ret = min(ret,query(mid+1,R,lt,rt,v<<1|1));
48     pushup(v);
49     return ret;
50 }
51 int main() {
52     int n,m;
53     while(scanf("%d%d",&n,&m),n||m) {
54         for(int i = 1; i <= n; ++i)
55             for(int j = 1; j <= m; ++j)
56                 scanf("%d",T[i] + j);
57         for(int i = 1; i <= n; ++i)
58             for(int j = 1; j <= m; ++j)
59                 scanf("%d",F[i] + j);
60         for(int i = 1; i <= m; ++i) dp[1][i] = T[1][i];
61         for(int i = 2; i <= n; ++i) {
62             build(1,m,1);
63             for(int j = 1; j <= m; ++j)
64                 update(1,m,max(1,j - F[i-1][j]),min(j + F[i-1][j],m),dp[i-1][j],1);
65             for(int j = 1; j <= m; ++j) {
66                 int tmp = query(1,m,max(1,j - F[i][j]),min(m,j + F[i][j]),1);
67                 dp[i][j] = min(INF,tmp + T[i][j]);
68             }
69         }
70         int ret = INF;
71         for(int i = 1; i <= m; ++i)
72             ret = min(ret,dp[n][i]);
73         printf("%d\n",ret);
74     }
75     return 0;
76 }