題目連結

思路：動態規劃，這裡下降的方程就是01揹包了，上升的由於到m之後還能繼續點保持m的高度，所以是完全揹包，但是由於同時只能執行一個，且若是先轉移01揹包的狀態會影響到後續完全揹包的轉移，所以優先轉移完全揹包，再轉移01揹包。 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<vector>
#include<queue>
#include<deque>
using namespace std;
#define ll long long
#define PI acos(-1)
#define INF 0x3f3f3f3f
#define NUM 1005
#define debug true
#define lowbit(x) ((-x)&x)
#define ffor(i,d,u) for(int i=(d);i<=(u);++i)
#define _ffor(i,u,d) for(int i=(u);i>=(d);--i)
#define mst(array,Num,Kind,Count) memset(array,Num,sizeof(Kind)*(Count))
const int P = 1e9+7;
int n,m,k;
int x[10005],y[10005],num = 0;
struct node
{
    int l,h;
}p[10005];
int dp[10005][1005];
int ans = INF;
bool flag;
template <typename T>
void read(T& x)
{
    x=0;
    char c;T t=1;
    while(((c=getchar())<'0'||c>'9')&&c!='-');
    if(c=='-'){t=-1;c=getchar();}
    do(x*=10)+=(c-'0');while((c=getchar())>='0'&&c<='9');
    x*=t;
}
template <typename T>
void write(T x)
{
    int len=0;char c[21];
    if(x<0)putchar('-'),x*=(-1);
    do{++len;c[len]=(x%10)+'0';}while(x/=10);
    _ffor(i,len,1)putchar(c[i]);
}
void AC()
{
    int pos;
    read(n),read(m),read(k);
    for(int i=0;i<n;++i)read(x[i]),read(y[i]);
    for(int i=0;i<=n;++i)p[i].l = 0,p[i].h = m+1;
    for(int i=1;i<=k;++i)
    {
        read(pos);
        read(p[pos].l),read(p[pos].h);
    }
    for(int i=1;i<=m;++i)dp[0][i] = 0;
    for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)dp[i][j] = INF;
    for(int i=0;i<n;++i)
    {
        flag = false;
        int k;
        for(int j=p[i].l+1;j<p[i].h;++j)//完全揹包
        {
            if(p[i+1].h > m&&j+x[i] > m)
                dp[i+1][m] = min(dp[i+1][m],dp[i][j]+1);
            else
                if(j+x[i] < p[i+1].h)
                {//計算從i,j跳到i+1最少要點選幾次螢幕
                    if(j+x[i] <= p[i+1].l)k = (p[i+1].l-j)/x[i]+1;
                    else k = 1;
                    dp[i+1][j+k*x[i]] = min(dp[i+1][j+k*x[i]],dp[i][j]+k);
                }
            if(dp[i][j] != INF&&p[i].h <= m)flag = true;
        }
        for(int j=p[i+1].l+1;j<p[i+1].h;++j)//01揹包
        {
            if(p[i+1].h > m&&j+x[i] > m)
                dp[i+1][m] = min(dp[i+1][m],dp[i+1][j]+1);
            else
                if(j+x[i] < p[i+1].h)
                    dp[i+1][j+x[i]] = min(dp[i+1][j+x[i]],dp[i+1][j]+1);
        }
        for(int j=p[i].l+1;j<p[i].h;++j)
            if(j-y[i] > p[i+1].l)dp[i+1][j-y[i]] = min(dp[i+1][j-y[i]],dp[i][j]);
        if(flag)++num;
    }
    for(int i=p[n].l+1;i<p[n].h;++i)ans=min(ans,dp[n][i]);
    if(ans != INF)printf("1\n%d",ans);
    else printf("0\n%d",num);
}
int main()
{
    AC();
    return 0;
}