经过自己的理解及找bug,写了不同含义的dp数组(每次结束时dp[i][j]包括当前次的c[i]),巩固了dp理解......
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 10;
ll dp[maxn][maxn];
const int INF = 0x3f3f3f3f;
ll a[maxn], b[maxn], c[maxn];
int N, C;
int main()
{
int T;
cin >> T;
while(T--)
{
scanf("%d%d", &N, &C);
for(int i = 1; i <= N; i++)
{
scanf("%d%d%d", &a[i], &b[i], &c[i]);
}
for(int i = 0; i <= N; i++)
{
for(int j = 0; j <= N; j++)
{
dp[i][j] = -INF;
}
}
dp[0][0] = C;
a[0] = 0;
b[0] = 0;
c[0] = 0;
for(int i = 1; i <= N; i++)
{
for(int j = 0; j <= i; j++)
{
if(j)
{
if(dp[i - 1][j - 1] >= 0)
{
int t = min(dp[i - 1][j - 1], b[i]);
if(t > a[i])
{
dp[i][j] = max(dp[i][j], t - a[i] + c[i]);
}
}
}
if(dp[i - 1][j] > 0) //必须加这个判断条件!不然<0的可能加上几次c[i]后>0,根据题意任意时刻体力不能为<=0,而这样把本不能继续挑战的变为能继续挑战,就错了(c[i]数据范围蛮大, 到1e9,很有可能累加后>0)
dp[i][j] = max(dp[i][j], dp[i - 1][j] + c[i]);
}
}
for(int i = N; i >= 0; i--)
{
if(dp[N][i] >= 0)
{
cout << i << endl;
break;
}
}
}
return 0;
}
/*
2
3 10
1 2 0
4 8 3
6 10 1
2 1
1 1 1
1 1 1
*/