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 = 110, M = 10010;
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 = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; 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, 0, sizeof(dp));
        ans = 0;
        if(n == 0 && m == 0 && k == 0) break;
        ++n; 
        ++m;
        for(int i = 1; i <= n; ++i)
            for(int j = 2; j <= m; ++j) 
            {
                int x;
                x = read();
                sum[i][j] = sum[i][j - 1] + x;
            }
        for(int i = 1; i <= n; ++i)
            for(int j = 2; j <= m; ++j) 
            {
                int x;
                x = read();
                dis[i][j] = dis[i][j - 1] + x;
            }
        for(int i = 1; i <= n; ++i)
        {
            int l = 1, r = 0;
            q[++r] = 1;
            for(int j = 1; j <= m; ++j) 
            {
                while(l <= r && dp[i - 1][j] - sum[i][j] > dp[i - 1][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 - 1][q[l]] + sum[i][j] - sum[i][q[l]];
            } 
            l = 1; 
            r = 0;
            q[++r] = m;
            for(int j = m; j; --j)
            {
                while(l <= r && dp[i - 1][j] + sum[i][j] > dp[i - 1][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 - 1][q[l]] + sum[i][q[l]] - sum[i][j]);
                if(i == n) ans = max(ans, dp[i][j]);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}