首页 > 其他 > 详细

P1941题解|NOIP2014飞扬的小鸟

时间:2018-01-21 15:16:50      阅读:202      评论:0      收藏:0      [点我收藏+]

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;
}

P1941题解|NOIP2014飞扬的小鸟

原文:https://www.cnblogs.com/iyanhang/p/8324233.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!