[题目链接]

         http://uoj.ac/problem/17

[算法]

        动态规划，f[i][j]表示横坐标为i，高度为j时，最少需要点击屏幕的次数，转移类似于背包

        时间复杂度 : O(NM)

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXN 10010
#define MAXM 1010
const int inf = 2e9;

int i,j,n,m,k,P,L,H,cnt,ans,t;
int X[MAXN],Y[MAXN],l[MAXN],r[MAXN];
int f[MAXN][MAXM];
 
namespace IO
{
        template <typename T> inline void read(T &x)
        {
                int f = 1; x = 0;
                char c = getchar();
                for (; !isdigit(c); c = getchar())
                {
                        if (c == '-') f = -f;
                }
                for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
                x *= f;
        }
        template <typename T> inline void write(T x)
        {
                if (x < 0)
                {
                        putchar('-');
                        x = -x;
                }
                if (x > 9) write(x / 10);
                putchar(x % 10 + '0');
        }
        template <typename T> inline void writeln(T x)
        {
                write(x);
                puts("");
        }
} ;

int main() 
{
        
        IO :: read(n); IO :: read(m); IO :: read(k);
        for (i = 0; i < n; i++)
        {
                IO :: read(X[i]);
                IO :: read(Y[i]);
        }
        for (i = 0; i <= n; i++)
        {
                l[i] = 0;
                r[i] = m + 1;
        }
        for (i = 1; i <= k; i++)
        {
                IO :: read(P); IO :: read(L); IO :: read(H);
                l[P] = L;
                r[P] = H;
        }
        for (i = 0; i <= n; i++)
        {
                for (j = 0; j <= m; j++)
                {
                        f[i][j] = inf;
                }
        }
        for (i = 1; i <= m; i++) f[0][i] = 0;
        for (i = 1; i <= n; i++)
        {
                for (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 (t = j - X[i - 1]; t <= m; t++) 
                                {
                                        f[i][m] = min(f[i][m],f[i - 1][t] + 1);
                                        f[i][m] = min(f[i][m],f[i][t] + 1);
                                }
                        }
                }    
                for (j = l[i] + 1; j <= r[i] - 1; j++) 
                {
                        if (j + Y[i - 1] <= m)
                                f[i][j] = min(f[i][j],f[i - 1][j + Y[i - 1]]);
                }
                for (j = 1; j <= l[i]; j++) f[i][j] = inf;
                for (j = r[i]; j <= m; j++) f[i][j] = inf;
        }
        cnt = k; ans = inf;
        for (i = n; i >= 1; i--)
        {
                for (j = l[i] + 1; j <= r[i] - 1; j++) ans = min(ans,f[i][j]);
                if (ans != inf) break;
                if (r[i] <= m) cnt--;        
        }
        if (cnt != k) 
        {
                IO :: writeln(0);
                IO :: writeln(cnt);
        } else
        {
                IO :: writeln(1);
                IO :: writeln(ans);
        }
        
        return 0;
    
}