uvalive4327（单调队列优化）

这题我有闪过是用单调队列优化的想法，也想过有左右两边各烧一遍。 但是不敢确定，搜了题解，发现真的是用单调队列，然后写了好久，调了好久下标应该怎么变化才过的。

dp[i][j] 表示走到第i行，第j个竖线的最大价值。

dp[i][j] = max(dp[i-1][k]+pre[i][j-1]-pre[i][k-1]);  从左往右

dp[i][j] = max(dp[i][j],dp[i-1][k]+suf[i][j]-suf[i][k]); 从右往左

 1 #pragma warning(disable:4996)
 2 #pragma comment(linker, "/STACK:1024000000,1024000000")
 3 #include <stdio.h>
 4 #include <string.h>
 5 #include <time.h>
 6 #include <math.h>
 7 #include <map>
 8 #include <set>
 9 #include <queue>
10 #include <stack>
11 #include <vector>
12 #include <bitset>
13 #include <algorithm>
14 #include <iostream>
15 #include <string>
16 #include <functional>
17 const int INF = 1 << 30;
18 typedef __int64 LL;
19 /*
20 dp[i][j]表示走到第i个横线，第j个竖线的最大值
21 dp[i][j] = max(dp[i-1][k] + sum[j-1] - sum[k-1])
22 单调队列优化维护队首最大值，
23 */
24
25 int q[11111], head, tail;
26 int dp[111][11111];
27 int a[111][11111];
28 int b[111][11111];
29 int pre[111][11111], suf[111][11111];
30 int sum1[111][11111];
31 int sum2[111][11111];
32 int main()
33 {
34     int n, m, k;
35
36     while (scanf("%d%d%d", &n, &m, &k),n+m+k)
37     {
38         memset(dp, 0, sizeof(dp));
39         memset(pre, 0, sizeof(pre));
40         memset(suf, 0, sizeof(suf));
41         memset(sum1, 0, sizeof(sum1));
42         memset(sum2, 0, sizeof(sum2));
43         n++;
44         for (int i = 1;i <= n;++i)
45         {
46             for (int j = 1;j <= m;++j)
47             {
48                 scanf("%d", &a[i][j]);
49                 pre[i][j] = pre[i][j - 1] + a[i][j];
50                 suf[i][j] = a[i][j];
51             }
52             for (int j = m;j >= 1;--j)
53                 suf[i][j] += suf[i][j + 1];
54
55         }
56         for (int i = 1;i <= n;++i)
57         {
58             for (int j = 1;j <= m;++j)
59             {
60                 scanf("%d", &b[i][j]);
61                 sum1[i][j] = b[i][j];
62                 sum2[i][j] = sum2[i][j - 1] + b[i][j];
63             }
64             for (int j = m;j >= 1;--j)
65                 sum1[i][j] += sum1[i][j + 1];
66         }
67         for (int i = n;i >= 1;--i)
68         {
69             head = tail = 0;
70             for (int j = 1;j <= m + 1;++j)
71             {
72                 while (head < tail && dp[i + 1][j] - pre[i][j - 1] >= dp[i + 1][q[tail - 1]] - pre[i][q[tail - 1] - 1])
73                     tail--;
74                 q[tail++] = j;
75                 while (head<tail && sum2[i][j-1] - sum2[i][q[head]-1]>k)
76                     head++;
77                 dp[i][j] = dp[i + 1][q[head]] + pre[i][j - 1] - pre[i][q[head]-1];
78             }
79             head = tail = 0;
80
81             for (int j = m+1;j >= 1;--j)
82             {
83                 while (head < tail&&dp[i + 1][j] - suf[i][j ] >= dp[i + 1][q[tail - 1]] - suf[i][q[tail - 1] ])
84                     tail--;
85                 q[tail++] = j;
86                 while (head<tail&&sum1[i][j] - sum1[i][q[head]]>k)
87                     head++;
88                 dp[i][j] =std::max(dp[i][j], dp[i + 1][q[head]] + suf[i][j] - suf[i][q[head]]);
89             }
90         }
91         int ans = 0;
92         for (int j = 1;j <= m + 1;++j)
93             ans = std::max(ans, dp[1][j]);
94         printf("%d\n", ans);
95     }
96     return 0;
97 }