UVA1336 Fixing the Great Wall

Fixing the Great Wall

https://odzkskevi.qnssl.com/b7d37f69f479d57e44735fc5d6403983?v=1508326275

【题解】

按照x排序

dp[i][j][0/1]表示从i到j全修好,当前在i/j的最小代价和已知的未修好的时间附加代价(td)
转移即可，即

dp[i][j][0/1] - > d[i-1][j][0]

dp[i][j][0/1] - > d[i][j + 1][1]

为什么这样？想想不这样的情况，会发现都不如这样优秀

还有一个小处理，把起点虚成一个节点做

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector> 
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))

inline void swap(int &a, int &b)
{
    long long tmp = a;a = b;b = tmp;
}

inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
}

const int INF = 0x3f3f3f3f;
const int MAXN = 10000 + 10;

int n, v, s, sum[MAXN];
struct Node
{
    int x,c,d;
}node[MAXN];

int cmp(Node a, Node b)
{
    return a.x < b.x;
}

//dp[i][j][0/1]表示修复i~j这段区间，当前在最左端（0）， 最右端（1）的
//最小花费 

double dp[MAXN][MAXN][2];

int main()
{
    while(scanf("%d %d %d", &n, &v, &s) != EOF && n && v && s)
    {
        for(register int i = 1;i <= n;++ i)
            read(node[i].x), read(node[i].c), read(node[i].d);
        node[n + 1].x = s;
        node[n + 1].c = 0;
        node[n + 1].d = 0;
        ++ n;
        std::sort(node + 1, node + 1 + n, cmp);
        for(register int i = 1;i <= n;++ i)
            sum[i] = sum[i - 1] + node[i].d;
        for(register int i = 1;i <= n;++ i)
            for(register int j = 1;j <= n;++ j)
            {
                if(i == j && node[i].x == s && node[i].c == 0 && node[i].d == 0) dp[i][j][0] = dp[i][j][1] = 0;
                else dp[i][j][0] = dp[i][j][1] = INF;
            }
        for(register int len = 1;len <= n;++ len)
        {
            for(register int i = 1;i <= n;++ i)
            {
                int j = i + len - 1;
                dp[i - 1][j][0] = min(dp[i - 1][j][0],
                                      min(dp[i][j][0] + (double)(node[i].x - node[i - 1].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[i - 1].c,
                                          dp[i][j][1] + (double)(node[j].x - node[i - 1].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[i - 1].c));
                dp[i][j + 1][1] = min(dp[i][j + 1][1], 
                                      min(dp[i][j][0] + (double)(node[j + 1].x - node[i].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[j + 1].c,
                                            dp[i][j][1] + (double)(node[j + 1].x - node[j].x)/v * (sum[i - 1] + sum[n] - sum[j]) + node[j + 1].c));
            }
        }
        printf("%d
", min((int)dp[1][n][0], (int)dp[1][n][1]));
    }
    return 0;
}