ZJOI2006 物流运输

Posted 2020-10-25 zcdhj

LOJ

DP+最短路

可以先用 \(Dijkstra\) 将每一段时间内的费用都先预处理出来， \([i,j]\) 时间段的费用表示为 \(Cost(i,j)\) ，然后再设 \(dp[i]\) 为第 \(i\) 天时的最优解。状态转移的时候枚举第几天的时候改变了路线。

\[dp[i]=\min_{0≤j<i} \{dp[j]+Cost(j+1,i)*(i-j)+k\}\]
\[dp[0]=-k\]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

const int max_n = 100 + 5;
const int max_m = 20 + 5;
const int max_e = max_m * max_m + 5;
const int inf = 0x3f3f3f3f;

int N, M, K, E, D, Tot;
int Dp[max_n], First_edge[max_m], P[max_m][max_n], Flag[max_m], Dis[max_m], C[max_n][max_n];

struct Edge
{
    int to, w, next_edge;
}Linker[max_e << 1];

struct Heap
{
    int cur;
    Heap(int x)
    {
        cur = x;
    }
    bool operator < (const Heap &x) const
    {
        return Dis[cur] > Dis[x.cur];
    }
};

std::priority_queue <Heap> Q;

inline int read()
{
    register int x = 0;
    register char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch))
    {
        x = (x << 1) + (x << 3) + ch - ‘0‘;
        ch = getchar();
    }
    return x;
}

inline void add_edge(int x, int y, int z)
{
    Linker[++Tot].to = y;
    Linker[Tot].w = z;
    Linker[Tot].next_edge = First_edge[x];
    First_edge[x] = Tot;
}

int Dijkstra()
{
    int from, to, w;
    memset(Dis, inf, sizeof(Dis));
    Dis[1] = 0;
    Q.push(Heap(1));
    while(!Q.empty())
    {
        from = Q.top().cur;
        Q.pop();
        for(int k = First_edge[from]; k; k = Linker[k].next_edge)
        {
            to = Linker[k].to;
            w = Linker[k].w;
            if(Dis[to] > Dis[from] + w && !Flag[to])
            {
                Dis[to] = Dis[from] + w;
                Q.push(Heap(to));
            }
        }
    }
    return Dis[M];
}

int main()
{
    int x, y, z;
    N = read();
    M = read();
    K = read();
    E = read();
    for(int i = 1; i <= E; ++i)
    {
        x = read();
        y = read();
        z = read();
        add_edge(x, y, z);
        add_edge(y, x, z);
    }
    D = read();
    for(int i = 1; i <= D; ++i)
    {
        x = read();
        y = read();
        z = read();
        for(int j = y; j <= z; ++j)
            P[x][j] = 1;
    }
    for(int i = 1; i <= N; ++i)
    {
        for(int j = i; j <= N; ++j)
        {
            memset(Flag, 0, sizeof(Flag));
            for(int k = 1; k <= M; ++k)
                for(int l = i; l <= j; ++l)
                    Flag[k] |= P[k][l];
            C[i][j] = Dijkstra();
        }
    }
    memset(Dp, inf, sizeof(Dp));
    Dp[0]= -K;
    for(int i = 1; i <= N; ++i)
        for(int j = 0; j < i; ++j)
            Dp[i] = std::min(Dp[i], (C[j + 1][i] == inf) ? inf : Dp[j] + C[j + 1][i] * (i - j) + K);
    printf("%d\n", Dp[N]);
    return 0;
}