$dp(i,j)$表示i~j这段还没运走时的状态,包括 运输了多少次,还剩多少空间
每次枚举运输左边还是右边转移
#include <bits/stdc++.h> #define rep(i, j, k) for (int i = int(j); i <= int(k); ++ i) #define dwn(i, j, k) for (int i = int(j); i >= int(k); -- i) using namespace std; typedef pair<int, int> P; const int N = 1e4 + 7; P dp[2][N]; int w[N]; int main() { int n, t; scanf("%d%d", &t, &n); rep(i, 1, n) scanf("%d", w + i); P tmp; auto Enlarge = [&](P &a, P &b) { if (a.first == -1) a = b; else { if (a.first > b.first || (a.first == b.first && a.second < b.second)) a = b; } }; for (int i = 1; i <= n; ++ i) dp[(n & 1)][i].first = -1; dp[(n & 1)][1] = P(0, 0); dwn(l, n, 1) { // i, j 之间还没消去 int cur = l & 1, next = cur ^ 1; // cout << l << ' ' << dp[cur][3].first << ' ' << dp[cur][3].second << ' '; for (int i = 1; i <= n; ++ i) dp[next][i].first = -1; for (int i = 1; i + l - 1 <= n; ++ i) if (dp[cur][i].first != -1) { int j = i + l - 1; // 选第i个 if (w[i] <= dp[cur][i].second) { tmp = P(dp[cur][i].first, dp[cur][i].second - w[i]); Enlarge(dp[next][i + 1], tmp); } else { tmp = P(dp[cur][i].first + 1, t - w[i]); Enlarge(dp[next][i + 1], tmp); } // 选第jge if (w[j] <= dp[cur][i].second) { tmp = P(dp[cur][i].first, dp[cur][i].second - w[j]); Enlarge(dp[next][i], tmp); } else { tmp = P(dp[cur][i].first + 1, t - w[j]); Enlarge(dp[next][i], tmp); } } } // cout << dp[0][3].first << ' ' << dp[0][3].second << ' '; int ans = 10000000; for (int i = 1; i <= n; ++ i) if (dp[0][i].first != -1) ans = min(ans, dp[0][i].first); printf("%d ", ans); } /* 4 5 1 1 3 1 2 */