[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 = ;char c = getchar();
     while(c > '' || c < '') c = getchar();
     while(c >= '' && c <= '') x = x *  + c - '', 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, , sizeof(dis));
     dis[] = ;
     q.push(), z[] = 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 - ][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[]) cost[l][r] = dis[m] * (r - l + );
     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 = ; i <= e; i ++)
     {
         a = read(), b = read(), c = read();
         add(a, b, c);
     }
     d = read();
     for(R i = ; i <= d; i ++)
     {
         a = read(), b = read(), c =read();
         for(R j = b; j <= c; j ++) sum[j][a] = ;
     }
     for(R i = ; i <= m; i ++)//枚举码头
         for(R j = ; j <= n; j ++)
             sum[j][i] += sum[j - ][i];
 }

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

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