[luoguP1772] [ZJOI2006]物流运输（DP + spfa）

传送门

预处理cost[i][j]表示从第i天到第j天起点到终点的最短距离

f[i]表示前i天到从起点到终点的最短距离

f[0] = -K

f[i] = min(f[i], f[j - 1] + cost[j][i] + K)

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#define N 1001
#define min(x, y) ((x) < (y) ? (x) : (y))

int n, m, K, e, d, cnt;
int f[N], cost[N][N];
bool b[N][N], flag[N], vis[N];
int head[N], to[N], next[N], val[N], dis[N];
//f[i]表示前i天从起点到终点的最小费用
//cost[i][j]表示从第i天到, b[i][j]表示第i个码头第j天是否不能用 

std::queue <int> q;

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void add(int x, int y, int z)
{
	to[cnt] = y;
	val[cnt] = z;
	next[cnt] = head[x];
	head[x] = cnt++;
}

inline int spfa()
{
	int i, u, v;
	memset(vis, 0, sizeof(vis));
	memset(dis, 127, sizeof(dis));
	q.push(1);
	dis[1] = 0;
	while(!q.empty())
	{
		u = q.front();
		q.pop();
		vis[u] = 0;
		for(i = head[u]; i ^ -1; i = next[i])
		{
			v = to[i];
			if(flag[v]) continue;
			if(dis[v] > dis[u] + val[i])
			{
				dis[v] = dis[u] + val[i];
				if(!vis[v])
				{
					q.push(v);
					vis[v] = 1;
				}
			}
		}
	}
	return dis[m];
}

int main()
{
	int i, j, k, l, x, y, z;
	n = read();
	m = read();
	K = read();
	e = read();
	memset(head, -1, sizeof(head));
	for(i = 1; i <= e; i++)
	{
		x = read();
		y = read();
		z = read();
		add(x, y, z);
		add(y, x, z);
	}
	d = read();
	for(i = 1; i <= d; i++)
	{
		z = read();
		x = read();
		y = read();
		for(j = x; j <= y; j++) b[z][j] = 1;
	}
	for(i = 1; i <= n; i++)
		for(j = i; j <= n; j++)
		{
			memset(flag, 0, sizeof(flag));
			for(k = 1; k <= m; k++)
				for(l = i; l <= j; l++)
					flag[k] |= b[k][l];
			cost[i][j] = spfa();
		}
	memset(f, 127, sizeof(f));
	f[0] = -K;
	for(i = 1; i <= n; i++)
		for(j = 1; j <= i; j++)
			if(cost[j][i] <= 1e9)
				f[i] = min(f[i], f[j - 1] + cost[j][i] * (i - j + 1) + K);
	printf("%d\n", f[n]);
	return 0;
}