HDU 6024 Building Shops[dp]

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题意：给定n个点（n<=3000），每个点有个wi，对于每个点，选择的代价为wi，不选择的代价为跟他距离最近的左边的点的距离。问最小代价为多少。

分析：dp[i]表示最后取的点为 i 时的最小代价，那么转移的话就是dp[i] = min(dp[i],dp[j]+sum[i - 1] - sum[j] - x[j]*(i-j-1) + w[i]);

其中，sum[i]为1-i的所有点的坐标之和，复杂度为O(n^2)，好像能通过斜率优化到O(n),目前不会，，回头再补。

以下是代码。

#include<iostream>#include<cstdio>#include<algorithm>#include<queue>#include<map>#include<set>#include<stack>#include<cstring>#include<string>#include<vector>#include<iomanip>//#include<unordered_set>//#include<unordered_map>#include<cmath>#include<list>#include<bitset>using namespace std;#define _____ ios::sync_with_stdio(false); cin.tie(0);#define ull unsigned long long#define ll long long#define lson l,mid,id<<1#define rson mid+1,r,id<<1|1typedef pair<int, int>pii;typedef pair<ll, ll>pll;typedef pair<double, double>pdd;const double eps = 1e-6;const int MAXN = 300005;const int MAXM = 5005;const ll LINF = 0x3f3f3f3f3f3f3f3f;const int INF = 0x3f3f3f3f;const double FINF = 1e18;const ll MOD = 1000000007;struct lx {    ll xi, ci;    friend bool operator<(const lx a, const lx b) {        return a.xi < b.xi;    }}tmp[3005];ll dp[3005];ll sum[3005];int main(){    int n;    while (scanf("%d", &n) != EOF)    {        memset(dp, LINF, sizeof(dp));        for (int i = 1; i <= n; ++i)scanf("%I64d%I64d", &tmp[i].xi, &tmp[i].ci);        sort(tmp + 1, tmp + n + 1); sum[1] = tmp[1].xi;        for (int i = 2; i <= n; ++i)sum[i] = sum[i - 1] + tmp[i].xi;        dp[1] = tmp[1].ci;        for (int i = 2; i <= n; ++i)        {            for (int j = 1; j < i; ++j)            {                dp[i] = min(dp[i], dp[j] + sum[i - 1] - sum[j] - (i - j - 1)*tmp[j].xi + tmp[i].ci);            }        }        ll ans = LINF;        //for (int i = 1; i <= n; i++)printf("%I64d\n", dp[i]);        for (int i = 1; i <= n; ++i)ans = min(ans, dp[i] + sum[n] - sum[i] - (n - i)*tmp[i].xi);        printf("%I64d\n", ans);    }}

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