题解说是二分答案和前缀和，然而我愣是没看到……
但是这就是线段树板子题啊！
区间修改维护最小值，如果当前修改后的区间最小值小于0的话就说明这个订单无法完全满足。
竟然TLE了一个点，开氧气过了。
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #include<cstdlib> 7 #include<cctype> 8 #include<stack> 9 #include<queue> 10 #include<vector> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 1e6 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), las = ' '; 25 while(!isdigit(ch)) las = ch, ch = getchar(); 26 while(isdigit(ch)) ans = (ans << 3) + (ans << 1) + ch - '0', ch = getchar(); 27 if(las == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) putchar('-'), x = -x; 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, m, a[maxn]; 38 39 int l[maxn << 2], r[maxn << 2], Min[maxn << 2], lzy[maxn << 2]; 40 void build(int L, int R, int now) 41 { 42 l[now] = L; r[now] = R; 43 if(L == R) {Min[now] = read(); return;} 44 int mid = (L + R) >> 1; 45 build(L, mid, now << 1); 46 build(mid + 1, R, now << 1 | 1); 47 Min[now] = min(Min[now << 1], Min[now << 1 | 1]); 48 } 49 void pushdown(int now) 50 { 51 if(lzy[now]) 52 { 53 Min[now << 1] += lzy[now]; 54 Min[now << 1 | 1] += lzy[now]; 55 lzy[now << 1] += lzy[now]; 56 lzy[now << 1 | 1] += lzy[now]; 57 lzy[now] = 0; 58 } 59 } 60 void update(int L, int R, int now, int d) 61 { 62 if(l[now] == L && r[now] == R) 63 { 64 Min[now] -=d; lzy[now] -= d; 65 return; 66 } 67 pushdown(now); 68 int mid = (l[now] + r[now]) >> 1; 69 if(R <= mid) update(L, R, now << 1, d); 70 else if(L > mid) update(L, R, now << 1 | 1, d); 71 else update(L, mid, now << 1, d), update(mid + 1, R, now << 1 | 1, d); 72 Min[now] = min(Min[now << 1], Min[now << 1 | 1]); 73 } 74 int query(int L, int R, int now) 75 { 76 if(L == l[now] && R == r[now]) return Min[now]; 77 pushdown(now); 78 int mid = (l[now] + r[now]) >> 1; 79 if(R <= mid) return query(L, R, now << 1); 80 else if(L > mid) return query(L, R, now << 1 | 1); 81 else return min(query(L, mid, now << 1), query(mid + 1, R, now << 1 | 1)); 82 } 83 84 int main() 85 { 86 n = read(); m = read(); 87 build(1, n, 1); 88 for(int i = 1; i <= m; ++i) 89 { 90 int d = read(), L = read(), R = read(); 91 update(L, R, 1, d); 92 if(query(L, R, 1) < 0) {printf("-1 %d ", i); return 0;} 93 } 94 write(0); enter; 95 return 0; 96 }