Petrozavodsk Winter Camp, Day 8, 2014, Ship

$dp(i,j)$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 
#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 << '\n';
        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 << '\n';
    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\n", ans);

}
/*
4 5
1 1 3 1 2
*/

 

转载于:https://www.cnblogs.com/tempestT/p/10673515.html