题意:用机器人修复分布在一条直线段上的n个损坏点,对于每个点i,x[i]表示位置,c[i]为立刻修缮的费用,d[i]为单位时间增加的维修费用。例如若在时刻t修缮i号点,则费用为ci+t∗di。给出机器人的初始位置x,速度v。求出修缮所有点的最小费用。
分析:在任意时刻,机器人修完的点肯定是一个连续的区间,这个区间的费用不会再增加,它接下来每移动一步,增加的费用都等于区间外的d之和乘以时间。f[i][j][k]为修复完i~j,当前位置在k(k为0则在x[i],为1则在x[j])之后所有的费用。可以用Color Length这道题的思路,即转移时每走一单位时间,就把这一单位时间中为修缮的点所产生的费用全部累加起来。(f[i][j][0/1]表示i到j这一段已经修完了,且机器人在左/右端,最小化费是多少,然后转移就行了 )
代码:
#include <bits/stdc++.h>
#define maxn 1010
#define ll long long
#define linf (1ll<<60)
using namespace std;
struct point
{
ll x, c, d;
} pt[maxn];
ll N, V, x0, ans, f[maxn][maxn][2], s[maxn];
bool operator<(point p1, point p2)
{
return p1.x<p2.x;
}
void init()
{
ll i;
for(i=1; i<=N; i++)scanf("%lld%lld%lld",&pt[i].x,&pt[i].c,&pt[i].d);
N++;
pt[N].x=x0, pt[N].c=0, pt[N].d=0;
sort(pt+1,pt+N+1);
for(i=1; i<=N; i++)s[i]=s[i-1]+pt[i].d;
}
void work()
{
ll i, j, l, ans;
for(i=1; i<=N; i++)
for(j=i; j<=N; j++)
if(i==j and pt[i].x==x0)f[i][j][0]=f[i][j][1]=0;///发酵点
else f[i][j][0]=f[i][j][1]=linf;
for(l=2; l<=N; l++) ///枚举区间长度
for(i=1; i+l-1<=N; i++)
{
j=i+l-1;
f[i][j][0]=min(f[i+1][j][0]+(pt[i+1].x-pt[i].x)*(s[N]-(s[j]-s[i])),f[i+1][j][1]+(pt[j].x-pt[i].x)*(s[N]-(s[j]-s[i])));
f[i][j][1]=min(f[i][j-1][0]+(pt[j].x-pt[i].x)*(s[N]-(s[j-1]-s[i-1])),f[i][j-1][1]+(pt[j].x-pt[j-1].x)*(s[N]-(s[j-1]-s[i-1])));
}
ans=min(f[1][N][0],f[1][N][1])/V;///速度延后除 简洁代码
for(i=1; i<=N; i++)ans+=pt[i].c;///费用延后加
printf("%lld\n",ans);
}
int main()
{
while((~scanf("%lld%lld%lld",&N,&V,&x0)) and (N+V+x0))init(), work();
return 0;
}
LRJ代码(记忆化搜索):
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1000 + 5;
const double INF = 1e30;
struct Section {
double x, c, dt;
bool operator < (const Section& rhs) const {
return x < rhs.x;
}
} s[maxn];
int kase, n;
int vis[maxn][maxn][2];
double v, x, d[maxn][maxn][2];
double psdt[maxn]; // prefix sum of dt
// cost accumulated when walking from x1 and x2.
// section[i~j] are already finished
double cost(double x1, double x2, int i, int j) {
double finished_dt = 0;
assert(i <= j);
if(i >= 0 && j >= 0) finished_dt += psdt[j] - psdt[i-1];
return (psdt[n] - finished_dt) * fabs(x2 - x1) / v;
}
double dp(int i, int j, int p) {
if(i == 1 && j == n) return 0;
double& ans = d[i][j][p];
if(vis[i][j][p] == kase) return ans;
vis[i][j][p] = kase;
ans = INF;
double x = (p == 0 ? s[i].x : s[j].x);
if(i > 1) ans = min(ans, dp(i-1, j, 0) + cost(x, s[i-1].x, i, j));
if(j < n) ans = min(ans, dp(i, j+1, 1) + cost(x, s[j+1].x, i, j));
return ans;
}
int main() {
memset(vis, 0, sizeof(vis));
while(scanf("%d%lf%lf", &n, &v, &x) == 3 && n) {
++kase;
double sumc = 0;
for(int i = 1; i <= n; i++) {
scanf("%lf%lf%lf", &s[i].x, &s[i].c, &s[i].dt);
sumc += s[i].c;
}
sort(s+1, s+n+1); // in increasing order of position
psdt[0] = 0;
for(int i = 1; i <= n; i++)
psdt[i] = psdt[i-1] + s[i].dt;
s[0].x = -INF;
s[n+1].x = INF;
double ans = INF;
for(int i = 1; i <= n+1; i++)
if(x > s[i-1].x && x < s[i].x) {
if(i > 1) ans = min(ans, dp(i-1, i-1, 0) + cost(x, s[i-1].x, -1, -1)); // move left
if(i <= n) ans = min(ans, dp(i, i, 0) + cost(x, s[i].x, -1, -1)); // move right
break;
}
printf("%.0lf\n", floor(ans + sumc));
}
return 0;
}