poj3926

dp+优化

很明显可以用单调队列优化。

记录下自己犯的sb错误：  数组开小，sum没搞清。。。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = , M = ;
int n, m, k, ans;
int q[M];
int dp[N][M], sum[N][M], dis[N][M];
inline int max(int x, int y)
{
    return x > y ? x : y;
}
inline int read()
{
    int x = , f = ; char c = getchar();
    while(c < '' || c > '') { if(c == '-') f = -; c = getchar(); }
    while(c >= '' && c <= '') { x = x *  + c - ''; c = getchar(); }
    return x * f;
}
int main()
{
    //dp[i][j] = dp[i - 1][x] + sum[j] - sum[x - 1]
    //dp[i][j] = dp[i - 1][x] + sum[x] - sum[j - 1]
    while(scanf("%d%d%d", &n, &m, &k))
    {
        memset(dp, , sizeof(dp));
        ans = ;
        if(n ==  && m ==  && k == ) break;
        ++n;
        ++m;
        for(int i = ; i <= n; ++i)
            for(int j = ; j <= m; ++j)
            {
                int x;
                x = read();
                sum[i][j] = sum[i][j - ] + x;
            }
        for(int i = ; i <= n; ++i)
            for(int j = ; j <= m; ++j)
            {
                int x;
                x = read();
                dis[i][j] = dis[i][j - ] + x;
            }
        for(int i = ; i <= n; ++i)
        {
            int l = , r = ;
            q[++r] = ;
            for(int j = ; j <= m; ++j)
            {
                while(l <= r && dp[i - ][j] - sum[i][j] > dp[i - ][q[r]] - sum[i][q[r]]) --r;
                q[++r] = j;
                while(l <= r && dis[i][j] - dis[i][q[l]] > k) ++l;
                dp[i][j] = dp[i - ][q[l]] + sum[i][j] - sum[i][q[l]];
            }
            l = ;
            r = ;
            q[++r] = m;
            for(int j = m; j; --j)
            {
                while(l <= r && dp[i - ][j] + sum[i][j] > dp[i - ][q[r]] + sum[i][q[r]]) --r;
                q[++r] = j;
                while(l <= r && dis[i][q[l]] - dis[i][j] > k) ++l;
                dp[i][j] = max(dp[i][j], dp[i - ][q[l]] + sum[i][q[l]] - sum[i][j]);
                if(i == n) ans = max(ans, dp[i][j]);
            }
        }
        printf("%d\n", ans);
    }
    return ;
}