[bzoj2037][Sdoi2008]Sue的小球

题目链接

算是一道比较明显的区间$dp$，但是状态不同往常。

$dp[0/1][i][j]$为拿完第$i$个和第$j$个范围内的所有球后停在左边/右边的最小亏损。

$dp[0][i][j] = min(dp[0][i][j], min(dp[0][i + 1][j] +亏损, dp[1][i + 1][j] + 亏损))$

$dp[1][i][j] = min(dp[1][i][j], min(dp[1][i][j - 1] + 亏损, dp[0][i][j - 1] +亏损))$

亏损的计算方式为速度的区间和*距离

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1000 + 10;
struct node {
    int x, y, v;
}a[maxn];
bool cmp(node a, node b) {
    return a.x < b.x;
}
int dp[2][maxn][maxn], c[maxn];
int qi(int l, int r) {
    return c[r] - c[l - 1];
}
int main() {
    int n, q;
    int sum = 0;
    memset(dp, 0x3f, sizeof(dp));
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i].x);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i].y), sum += a[i].y;
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i].v);
    a[++n] = { q,0,0 };
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; i++)
        c[i] = c[i - 1] + a[i].v;
    for (int i = 1; i <= n; i++)
        if (a[i].x == q && a[i].v == 0)
            dp[0][i][i] = dp[1][i][i] = 0;
    for (int k = 2; k <= n; k++) {
        for (int i = 1; i + k - 1 <= n; i++) {
            int j = i + k - 1;
            dp[0][i][j] = min(dp[0][i][j], min(dp[0][i + 1][j] + (qi(1, i) + qi(j + 1, n)) * (a[i + 1].x - a[i].x), dp[1][i + 1][j] + (qi(1, i) + qi(j + 1, n)) * (a[j].x - a[i].x)));
            dp[1][i][j] = min(dp[1][i][j], min(dp[1][i][j - 1] + (qi(1, i - 1) + qi(j, n)) * (a[j].x - a[j - 1].x), dp[0][i][j - 1] + (qi(1, i - 1) + qi(j, n)) * (a[j].x - a[i].x)));
        }
    }
    printf("%.3f\n", (sum - min(dp[0][1][n], dp[1][1][n])) / 1000.0);
}