题解 P1941 飞翔的小鸟

Posted 2022-05-12 nhdr233

数据范围$n\leqslant 10,000, m \leqslant 1,000$，写$O(nm)$的背包完全是可以通过本题的。对于上升是一个完全背包，对于下降是一个01背包，则有

$$f[i][j] = min(f[i-1][j-x[i-1]]+1, f[i][j-x[i-1]]+1, f[i-1][j+y[i-1]])$$

如何实现别的题解说明的很清楚我也不再赘述

同学都知道一般背包是有压维写法的，数组大小开到$10,000*1,000$怎么说都有点勉强~~并不勉强只是想优化~~，来考虑一下压维要实现的一些细节，压维以后的方程式是这样的

$$f[j] = min(f[j-x[i-1]]+1, f[j+y[i-1]])$$

而对于$f[j-x[i-1]]$实际上包含的是两个状态,分别是$f[i-1][j-x[i-1]]$和$f[i][j-x[i-1]]$所以可以新建两个辅助数组，分别存下两个状态，最后结果要在两个中间取

至于枚举顺序就像完全背包一样从小到大枚举就行枚举到$j$时$f[j+y[i-1]]$存的状态$f[i-1][j+y[i-1]]$所以并不需要拿数组暂存

$$f_1[j] = min(f_1[j-x[i-1]]+1, f[j-x[i-1]]+1)$$

$$f_2[j] = f[j+y[i-1]]$$

$$f[j]=min(f_1[j], f_2[j])$$

对于$f_1$表示该高度由跳跃而来的最小步数,对于$f_2$表示由降落而来的最小步数

这样就成功的把时间复杂度为$O(nm)$的空间复杂度优化为$O(m)$

```

//2019/2/2->NHDR233->AtHM->luoguP1941
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<time.h>
using namespace std;
inline int in()
    int x = 0, ch = getchar();
    while (!isdigit(ch)) ch = getchar();
    while (isdigit(ch)) x = x*10+ch-‘0‘, ch = getchar();
    return x;

void input();
void smin(int& x, int y)
    if (x > y) x = y;


const int TOP = 10100, INF = 1e9;
int n, m, k;
int x[TOP], y[TOP<<1], ans[TOP<<1], tem1[TOP], tem2[TOP];
struct node
    int where, down, top;
s[TOP];
bool cmp(node a, node b)
    return a.where < b.where;


void work()
    sort(s+1, s+k+1, cmp);
    int now = 1;
    for (int i = m+1; i <= 2*m; ++i) ans[i] = INF;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= 2*m; ++j)
            tem1[j] = ans[j];
        bool flag = false;
        for (int j = 1; j <= m; ++j)
            tem2[j] = INF;
            if (j > x[i-1])
                smin(tem2[j], tem1[j-x[i-1]]+1);
                smin(tem2[j], tem2[j-x[i-1]]+1);
           
            ans[j] = min(tem2[j], tem1[j+y[i-1]]);
       
        for (int j = 0; j <= x[i-1]; ++j)
            smin(ans[m], tem1[m-j]+1);
            smin(ans[m], tem2[m-j]+1);
        
        if (s[now].where == i)
            for (int j = 1; j <= m; ++j)
                if (j > s[now].down and j < s[now].top) continue;
                ans[j] = INF;
           
            now++;
       
        for (int j = 1; j <= m; ++j)
            if (ans[j] < INF) flag = true;
       
        if (!flag)
            printf("0\n%d", now-2);
            return;
       
   
    int final = INF;
    for (int i = 1; i <= m; ++i)
        smin(final, ans[i]);
   
    printf("1\n%d", final);

int main()
    input();
    work();

void input()
    n = in(); m = in(); k = in();
    for (int i = 0; i < n; ++i)
        x[i] = in(); y[i] = in();
   
    for (int i = 1; i <= k; ++i)
        s[i].where = in();
        s[i].down = in();
        s[i].top = in();
   


```