[NOIP2014]飞扬的小鸟

[NOIP2014]飞扬的小鸟

时间限制：1 s 内存限制：128 MB
【题目描述】

这道题很容易想出一个O(NM^2)的动规
即f[i][j]表示飞到i列j高度位置需要的最小点击数，则

f[i][j]=min{f[i−1][j−k∗x]+k},1≤k≤j/x（O(M)）（不考虑掉下来的情况）

这显然会超时。

优化：
f[i][j]=min{f[i−1][j−k∗x]+k},1≤k≤j/x

f[i][j−x]=min{f[i−1][(j−x)−(k−1)∗x]+k−1},2≤k≤j/x

对比以上式子，我们可以发现当k≥2 时
f[i][j]=f[i][j−x]+1

于是我们得到了崭新的状态转移方程：
f[i][j]=min{f[i−1][j−x],f[i][j−x]}+1（O(1)）（不考虑掉下来的情况）

O(M)地处理完一列后再O(M)地扫一遍，检查掉下来更优的情况。
总复杂度O(NM)

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#define maxn  10010using namespace std;const int inf = 0x7ffffff;int n,m,k,p,l,h;int x[maxn],y[maxn],down[maxn], up[maxn];int f[maxn][1001];int main() {    freopen("bird.in","r",stdin);    freopen("bird.out","w",stdout);    scanf("%d%d%d",&n,&m,&k);    for (int i = 0; i < n; ++i)        scanf("%d %d", &x[i], &y[i]);    for (int i = 1; i <=n; ++i) {        down[i] = 0;        up[i] = m + 1;    }    for(int i = 1; i <= k; ++i) {        cin >> p >> l >> h;        down[p] = l;        up[p] = h;    }    for (int i = 1; i <= n; ++i)        for (int j = 0; j <= m; ++j)                 f[i][j] = inf;    f[0][0] = inf;    int arrive = k;    for (int i = 1; i <= n; ++i) {        for (int j = 1; j <= m; ++j) {            if(j >= x[i-1]){                f[i][j] = min(f[i][j], f[i-1][j-x[i-1]] + 1);                f[i][j] = min(f[i][j], f[i][j-x[i-1]] + 1);            }            if(j == m) {                for(int k=j-x[i-1];k<=m;k++) {                    f[i][j] = min(f[i][j], f[i-1][k] + 1);                    f[i][j] = min(f[i][j], f[i][k] + 1);                }            }        }        for (int j = down[i]+1; j <= up[i]-1; ++j)            if( j + y[i-1] <= m)                f[i][j] = min(f[i][j], f[i-1][j+y[i-1]]);        for (int j = 1; j <= down[i]; ++j) f[i][j] = inf;        for (int j = up[i]; j <= m; ++j) f[i][j] = inf;    }    int cnt = k, ans = inf;    for (int i = n; i >= 1; i--) {        for (int j = down[i]+1; j <= up[i]-1; ++j)            if (f[i][j] < inf)               ans = min(ans, f[i][j]);        if (ans != inf) break;        if (up[i] <= m)           cnt --;    }    if(cnt==k)        printf("1\n%d\n", ans);    else         printf("0\n%d\n", cnt);    return 0;}