P1941题解|NOIP2014飞扬的小鸟

洛谷P1941

#include <bits/stdc++.h>
using namespace std;

const int inf=1e9;
const int MN=10001;
const int MM=1001;
const int MK=10001;

struct tube
{int p,l/*downward*/,h/*upward*/;}s[MK];
bool cmp(tube a,tube b){return a.p<b.p;}

int n,m,k,cnt=1;
int up[MN],down[MN];
int f[MN][MM];

int dp()
{
	for (int i=1;i<=n;++i)
	{
		for (int j=1;j<=m;++j) f[i][j]=inf;

		for (int j=1;j<=m;++j)
			if (j-up[i]>0)
				f[i][j]=min(f[i][j],min(f[i-1][j-up[i]],f[i][j-up[i]])+1);
		for (int j=m-up[i]+1;j<=m;++j) //'=' must be concentrated,for f[i-1][m] must be calced.don't for f[i][m] is twice and cut the '='. 
			f[i][m]=min(f[i][m],min(f[i][j],f[i-1][j])+1); //!!!

		for (int j=1;j<=(m-down[i]);++j)
			f[i][j]=min(f[i][j],f[i-1][j+down[i]]);

		if (i==s[cnt].p)
		{
			for (int j=1;j<=m;++j)
				if (j>=s[cnt].h || j<=s[cnt].l) f[i][j]=inf;
			bool flag=true;
			for (int j=1;j<=m;++j)
				if (f[i][j]!=inf)
				{
					flag=false;
					break;
				}
			if (flag) {cout<<"0\n"<<cnt-1; return 0;}
			cnt++;
		}
	}
	int ans=inf;
	for (int i=1;i<=m;++i) ans=min(ans,f[n][i]);
	cout<<"1\n"<<ans;
	return 0;
}
int init()
{
	cin>>n>>m>>k;
	for (int i=1;i<=n;++i)
		cin>>up[i]>>down[i];
	for (int i=1;i<=k;++i)
		cin>>s[i].p>>s[i].l>>s[i].h;
	sort(s+1,s+k+1,cmp);
	
	return 0;
}
int main()
{
	init();
	dp();

	return 0;
}