飞扬的小鸟

做法：dp

思路：

• 1.先用结构体把每个横坐标的情况记录下来
• 2.dp[i][j]表示到坐标(i,j)最少需要跳的次数
• 3.考虑上升转移和下降转移，其中上升转移要考虑到顶这种特殊情况

代码

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=10010,M=1010;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);

int n,m,k;
struct node{
    int x,y,minh,maxh;
    bool st;
}a[N];
int dp[N][M<<1]; //dp[i][j]:到坐标(i,j)最少需要跳的次数
int ans=INF;

void solve(){
    cin>>n>>m>>k;
    rep(i,1,n) {
        cin>>a[i].x>>a[i].y;
        a[i].minh=0,a[i].maxh=m+1;
    }
    while(k--){
        int p,l,h;
        cin>>p>>l>>h;
        a[p].minh=l,a[p].maxh=h,a[p].st=1;
    }
    mst(dp,INF);
    rep(i,1,m) dp[0][i]=0;
    rep(i,1,n){
        rep(j,1,m) dp[i][j+a[i].x]=min(dp[i-1][j]+1,dp[i][j]+1);
        rep(j,m+1,m+a[i].x) dp[i][m]=min(dp[i][j],dp[i][m]);//到顶 
        rep(j,1,m-a[i].y) dp[i][j]=min(dp[i][j],dp[i-1][j+a[i].y]);
        rep(j,1,a[i].minh) dp[i][j]=INF;
        rep(j,a[i].maxh,m) dp[i][j]=INF;
    }
    rep(i,1,m) ans=min(ans,dp[n][i]);
    if(ans<INF){
        cout<<"1\n"<<ans<<"\n";
        return;
    }
    cout<<"0\n";
    int i,j,num=0;
    for(i=1;i<=n;i++){
        for(j=1;j<=m;j++){
            if(dp[i][j]<INF) break;
        }
        if(j==m+1) break;
        if(a[i].st) num++;
    }
    cout<<num<<"\n";
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
#ifdef DEBUG
    freopen("F:/laji/1.in", "r", stdin);
//    freopen("F:/laji/2.out", "w", stdout);
#endif
//    int t;cin>>t;while(t--)
    solve();
    return 0;
}