[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  10010
using 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;
}