UVA LA 3983 - Robotruck DP，优先队列 难度: 2

题目

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1984

题意

按顺序给出平面n个点，每个点上都有重物，一次只能按顺序拿重量不超过c的重物，从原点出发并且返回原点，问至少多少次能把全部重物拿回原点

思路

明显，令dp[i]为从1拿到i所需的最小出发次数，令dist0[i]为从原点到i的距离，令dist[i]为从i到i-1的距离（点0不妨设置为原点），令distSum[i] = sum(dist[0]...dist[i])，则dp[j] = dp[j] + dist0[j] + distSum[j] + dp[i] + dist0[i + 1] - distSum[i+1]，此处j > i且i+1...j的重物重量和不超过c。设mycost[i] = dp[i] + dist0[i + 1] - distSum[i+1]，明显，可以使用尺取+优先队列来维护重量和不超过c和 dp[i] + dist0[i + 1] - distSum[i+1]最小这两个条件。

但刘书提出了进一步优化：由于当新加入的点i的mycost比尺取区间中维护的旧点小的时候，这些旧点是没有用的，可以不考虑它们。当这样做了之后，剩下的点总能成为一个递增序列，这样就非常容易维护。

感想

一开始DP公式推错了，弄成了还从该点出发

代码

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <tuple>
#define LOCAL_DEBUG
using namespace std;
typedef pair<int, int> MyPair;
const int MAXN = 1e5 + ;
int n, c;
int x[MAXN];
int y[MAXN];
int w[MAXN];
int wSum[MAXN];
int distSum[MAXN];
int dist0[MAXN];
int dp[MAXN];
int mydeque[MAXN];
int mycost[MAXN];

#define MYCOST(x) (dist0[(x) + 1] - distSum[(x) + 1] + dp[(x)])

int main() {
#ifdef LOCAL_DEBUG
    freopen("C:\\Users\\Iris\\source\\repos\\ACM\\ACM\\input.txt", "r", stdin);
    //freopen("C:\\Users\\Iris\\source\\repos\\ACM\\ACM\\output.txt", "w", stdout);
#endif // LOCAL_DEBUG
    int T;
    scanf("%d", &T);
    for (int ti = ; ti <= T && scanf("%d%d", &c, &n) == ; ti++) {
        for (int i = ; i <= n; i++) {
            scanf("%d%d%d", x + i, y + i, w + i);
            dist0[i] = abs(x[i]) + abs(y[i]);
        }
        for (int i = ; i <= n; i++) {
            wSum[i] = w[i] + wSum[i - ];
            distSum[i] = abs(x[i] - x[i - ]) + abs(y[i] - y[i - ]);
            distSum[i] += distSum[i - ];
        }

        int dequeTop = ;
        int dequeEnd = ;
        mydeque[dequeEnd++] = ;
        for (int i = ; i <= n; i++) {
            while (dequeTop < dequeEnd && wSum[i] - wSum[mydeque[dequeTop]] > c) {
                dequeTop++;
            }
            assert(dequeTop < dequeEnd);
            int f = mydeque[dequeTop];
            dp[i] = dist0[i] + dist0[f + ] + distSum[i] - distSum[f + ] + dp[f];
            while (dequeTop < dequeEnd && MYCOST(mydeque[dequeEnd - ]) >= MYCOST(i)) {
                dequeEnd--;
            }
            mydeque[dequeEnd++] = i;
        }
        if (ti != )puts("");
        printf("%d\n", dp[n]);
    }

    return ;
}