[Bzoj1003][ZJOI2006]物流运输(spfa+dp)

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=1003

比较简单的dp，dp[i]为1-i天最小费用，dp方程为dp[i] = min(dp[i], dp[j] + c[j + 1][i] * (i - j) + k),(0<=j<i)，c[i][j]为第i天到第j天都能走的最短路。

因为数据贼小，所有求最短路的方法都可以预处理出来c数组。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int INF = 1e9 + ;
 struct node {
     int e, w, next;
 }edge[];
 int head[], len;
 ll dp[];
 int st[][];
 ll c[][];
 int dis[];
 int vis[], vis2[];
 void init() {
     memset(head, -, sizeof(head));
     len = ;
     memset(dp, 0x7f, sizeof(dp));
 }
 void add(int s, int e, int w) {
     edge[len].e = e;
     edge[len].w = w;
     edge[len].next = head[s];
     head[s] = len++;
 }
 int spfa(int n) {
     queue<int>q;
     for (int i = ; i <= n; i++)
         dis[i] = INF, vis2[i] = ;
     dis[] = ;
     vis2[] = ;
     q.push();
     while (!q.empty()) {
         int x = q.front();
         q.pop();
         vis2[x] = ;
         for (int i = head[x]; i != -; i = edge[i].next) {
             int y = edge[i].e;
             if (vis[y])continue;
             if (dis[y] > dis[x] + edge[i].w) {
                 dis[y] = dis[x] + edge[i].w;
                 if (!vis2[y]) {
                     q.push(y);
                     vis2[y] = ;
                 }
             }
         }
     }
     return dis[n];
 }
 int main() {
     int n, m, k, s, x, y, z;
     scanf("%d%d%d%d", &n, &m, &k, &s);
     init();
     for (int i = ; i <= s; i++) {
         scanf("%d%d%d", &x, &y, &z);
         add(x, y, z);
         add(y, x, z);
     }
     scanf("%d", &s);
     for (int i = ; i <= s; i++) {
         scanf("%d%d%d", &x, &y, &z);
         for (int j = y; j <= z; j++)
             st[x][j] = ;
     }
     for (int i = ; i <= n; i++) {
         for (int j = ; j <= n; j++) {
             memset(vis, , sizeof(vis));
             for (int q = i; q <= j; q++)
                 for (int w = ; w <= m; w++)
                     if (st[w][q])vis[w] = ;
             c[i][j] = spfa(m);
         }
     }
     for (int i = ; i <= n; i++) {
         dp[i] = (ll)c[][i] * i;
         for (int j = ; j < i; j++)
             dp[i] = min(dp[i], dp[j] + c[j + ][i] * (i - j) + k);
     }
     printf("%lld\n", dp[n]);
     return ;
 }