UVa 1336 Fixing the Great Wall (区间DP)

题意：给定 n 个结点，表示要修复的点，然后机器人每秒以 v 的速度移动，初始位置在 x，然后修复结点时不花费时间，但是如果有的结点暂时没修复，

那么每秒它的费用都会增加 d，修复要花费 c，坐标是 pos，问你最少花费是多少。

析：dp[i][j][k] 表示已经修复了 i-j 区间，并且当前在 k，那么两种方案，向左移动，或者向右移动，最后输出就好了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define print(a) printf("%d\n", (a))
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
    int pos, cost, det;
    bool operator < (const Node &p) const{
        return pos < p.pos;
    }
};
Node a[maxn];
double sum[maxn], v;
double dp[maxn][maxn][2];

double cal(int i, int j, int l, int r){
     double t = 1.0 * fabs(a[l].pos-a[r].pos) / v;
     double ans = sum[i-1] + sum[n+1] - sum[j];
     return ans * t;
}

int solve(){
    for(int i = 0; i <= n+1; ++i)
        for(int j = 0; j <= n+1; ++j)
            dp[i][j][0] = dp[i][j][1] = inf;

    int p = lower_bound(a+1, a+n+1, (Node){m, 0, 0}) - a;
    dp[p][p][0] = dp[p][p][1] = 0.0;

    for(int i = p; i > 0; --i){
        for(int j = p; j <= n+1; ++j){
            dp[i-1][j][0] = min(dp[i-1][j][0], dp[i][j][0]+cal(i, j, i, i-1)+a[i-1].cost);
            dp[i-1][j][0] = min(dp[i-1][j][0], dp[i][j][1]+cal(i, j, j, i-1)+a[i-1].cost);
            dp[i][j+1][1] = min(dp[i][j+1][1], dp[i][j][0]+cal(i, j, i, j+1)+a[j+1].cost);
            dp[i][j+1][1] = min(dp[i][j+1][1], dp[i][j][1]+cal(i, j, j, j+1)+a[j+1].cost);
        }
    }
    return min(dp[1][n+1][0], dp[1][n+1][1]);
}

int main(){
    while(scanf("%d %lf %d", &n, &v, &m) == 3 && n+v+m){
        for(int i = 1; i <= n; ++i)  scanf("%d %d %d", &a[i].pos, &a[i].cost, &a[i].det);
        a[n+1].pos = m;  a[n+1].cost = a[n+1].det = 0;
        sort(a+1, a+n+2);
        sum[0] = 0;
        for(int i = 1; i <= n+1; ++i) sum[i] = sum[i-1] + a[i].det;

        int ans = solve();
        printf("%d\n", ans);
    }
    return 0;
}