[ZJOI2006]物流运输 DP 最短路

～～～题面～～～

题解：

　　设f[i]表示到第i天的代价，cost[i][j]表示第i天到第j天采取同一种方案的最小代价。那么转移就很明显了，直接$n^2$枚举即可。

　　所以问题就变成了怎么获取cost数组。因为i到j都采取同一种方案，因此这种方案不能经过在i到j这些天出现故障的码头，因为要求的是最小代价，因此直接跑最短路，然后注意判断一下不能经过在i到j会出现故障的码头即可。

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 110
#define ac 450
int n, m, k, e, d;
int Head[AC], date[ac], Next[ac], len[ac], tot;
int dis[AC], f[AC], cost[AC][AC], sum[AC][AC];//sum[天][码头]
bool z[AC];

struct cmp{
    bool operator () (int a, int b)
    {
        return dis[a] < dis[b];
    }
};

priority_queue<int, vector<int>, cmp> q;

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

inline void add(int f, int w, int S)
{
    date[++tot] = w, Next[tot] = Head[f], Head[f] = tot, len[tot] = S;
    date[++tot] = f, Next[tot] = Head[w], Head[w] = tot, len[tot] = S;
}

void upmin(int &a, int b)
{
    if(b < a) a = b;
}

void spfa(int l, int r)
{
    int x, now;
    memset(dis, 127, sizeof(dis));
    dis[1] = 0;
    q.push(1), z[1] = true;
    while(!q.empty())
    {
        x = q.top();
        q.pop();
        z[x] = false;
        for(R i = Head[x]; i; i = Next[i])
        {
            now = date[i];
            if(sum[r][now] - sum[l - 1][now]) continue;//如果这几天中故障过就不能用
            if(dis[now] > dis[x] + len[i])
            {
                dis[now] = dis[x] + len[i];
                if(!z[now]) q.push(now), z[now] = true;
            }
        }
    }
    if(dis[m] != dis[0]) cost[l][r] = dis[m] * (r - l + 1);
    else cost[l][r] = dis[m];//防爆
}

void pre()
{
    int a, b, c;
    n = read(), m = read(), k = read(), e = read();
    for(R i = 1; i <= e; i ++)
    {
        a = read(), b = read(), c = read();
        add(a, b, c);
    }
    d = read();
    for(R i = 1; i <= d; i ++)
    {
        a = read(), b = read(), c =read();
        for(R j = b; j <= c; j ++) sum[j][a] = 1;
    }
    for(R i = 1; i <= m; i ++)//枚举码头
        for(R j = 1; j <= n; j ++)
            sum[j][i] += sum[j - 1][i];
}

void work()
{
    for(R i = 1; i <= n; i ++) 
        for(R j = i; j <= n; j ++) spfa(i, j);
    memset(f, 127, sizeof(f));
    f[0] = -k;//因为第一天不用承受换方案的代价
    for(R i = 1; i <= n; i ++)//枚举天数
        for(R j = 0; j < i; j ++)//枚举上一段的结尾
            upmin(f[i], f[j] + cost[j + 1][i] + k);
    printf("%d
", f[n]);
}

int main()
{
//    freopen("in.in", "r", stdin);
    pre();
    work();
//    fclose(stdin);
    return 0;
}