//第一次做区间DP的题 看了别人的题解写的
//解法只要将各个点(包括餐厅的位置)按照距离原点的距离升序排序
//然后从餐厅这个点开始向两侧区间DP
// dp[i][j][0] 表示 当前点送完了区间[i,j]的外卖并且在区间的左端点I点
// dp[i][j][1] 表示 当前点送完了区间[i,j]的外卖并且在区间的右端点J点
//那么dp[i][j][0] 可由dp[i+1][j][0] 和 dp[i+1][j][1] 推出
// dp[i][j][1] 可由dp[i][j-1][0] 和 dp[i][j-1][1] 推出
#include <iostream>
#include "stdio.h"
#include "stdlib.h"
#include "string.h"
#include "algorithm"
#include <queue>
#include <stack>
#define N 100005
#define INF 1<<30
using namespace std;
int dp[1005][1005][2];
int sum[1005];
struct node {
int x, v;
}point[1005];
int cmp(node a, node b)
{
return a.x < b.x;
}
int cal(int a, int b)
{
return sum[b]-sum[a - 1];
}
int main()
{
int n, v, x, op;
//freopen("t", "r", stdin);
while(scanf("%d%d%d", &n, &v, &x) != EOF)
{
for(int i = 1; i <= n; i++)
scanf("%d%d", &point[i].x, &point[i].v);
point[++n].x = x, point[n].v = 0;
sort(point+1, point+n+1, cmp);
sum[0] = 0;
for(int i = 1; i <= n; i++)
{
if(point[i].x == x )
{
op = i;
}
sum[i] = sum[i-1] + point[i].v;
}
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
dp[i][j][0] = dp[i][j][1] = INF;
dp[op][op][0] = dp[op][op][1] = 0;
for( int i = op; i >= 1; i--)
for( int j = op; j <= n; j++)
{
int cost = cal(1, i - 1) + cal(j + 1, n);
if(i == j) continue;
dp[i][j][0] = min( dp[i][j][0] , dp[i + 1][j][0] + (point[i+1].x - point[i].x) * ( cost + point[i].v) );
dp[i][j][0] = min( dp[i][j][0] , dp[i + 1][j][1] + (point[j].x - point[i].x) * ( cost +point[i].v ) );
dp[i][j][1] = min( dp[i][j][1] , dp[i][j - 1][0] + (point[j].x - point[i].x) * (cost + point[j].v ) );
dp[i][j][1] = min( dp[i][j][1] , dp[i][j - 1][1] + (point[j].x - point[j-1].x) * (cost + point[j].v ) );
}
printf("%d\n", v*(min( dp[1][n][0], dp[1][n][1])));
}
}
ZOJ 3469(区间DP)
最新推荐文章于 2020-08-10 20:38:05 发布