【BZOJ1003】[ZJOI2006]物流运输

【BZOJ1003】[ZJOI2006]物流运输

题面

洛谷

bzoj

题解

设(f_i)表示前(i)天花费的最小值。

我们设第(l,r)天(1)到(m)的距离为(dis_{l,r})，这个可以(n^2)遍最短路求出。

那么转移就很显然了：

[f_i=min_{j=0}^{i-1} f_j+dis_{j+1,i} imes (i-j)+K ]

注意初值(f_0=-K)，因为第一次更改不算贡献。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
#include <cmath> 
#include <algorithm>
#include <queue> 
using namespace std; 
const int MAX_N = 1e3 + 5; 
struct Graph { int to, cost, next; } e[MAX_N << 1]; 
int fir[MAX_N], e_cnt; 
void Add_Edge(int u, int v, int w) {
	e[e_cnt] = (Graph){v, w, fir[u]};
	fir[u] = e_cnt++; 
}
const int INF = 1e9; 
int N, M, K, E, D; 
bool run[MAX_N][MAX_N], ok[MAX_N], inq[MAX_N];
int dis[MAX_N][MAX_N], ds[MAX_N]; 
void spfa(int l, int r) {
	for (int i = 1; i <= M; i++) {
		ok[i] = 1;
		for (int j = l; j <= r && ok[i]; j++) if (run[i][j]) ok[i] = 0; 
	} 
	queue<int> que;
	for (int i = 1; i <= M; i++) ds[i] = INF, inq[i] = 0; 
	ds[1] = 0, inq[1] = 1, que.push(1); 
	while (!que.empty()) { 
		int x = que.front(); que.pop(); 
		for (int i = fir[x]; ~i; i = e[i].next) {
			int v = e[i].to; if (!ok[v]) continue; 
			if (ds[x] + e[i].cost < ds[v]) {
				ds[v] = ds[x] + e[i].cost;
				if (!inq[v]) que.push(v), inq[v] = 1; 
			} 
		} 
		inq[x] = 0; 
	}
	dis[l][r] = ds[M]; 
}
long long f[MAX_N]; 
int main () { 
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif
	memset(fir, -1, sizeof(fir)); 
	scanf("%d%d%d%d", &N, &M, &K, &E); 
	for (int i = 1; i <= E; i++) {
		int u, v, w; scanf("%d%d%d", &u, &v, &w); 
		Add_Edge(u, v, w), Add_Edge(v, u, w); 
	} 
	scanf("%d", &D); 
	for (int i = 1; i <= D; i++) {
		int p, l, r; scanf("%d%d%d", &p, &l, &r); 
		for (int j = l; j <= r; j++) run[p][j] = 1; 
	} 
	for (int i = 1; i <= N; i++)
		for (int j = i; j <= N; j++)
			spfa(i, j);
	f[0] = -K; 
	for (int i = 1; i <= N; i++) { 
		f[i] = 1e18;
		for (int j = 0; j < i; j++)
			f[i] = min(f[i], f[j] + 1ll * dis[j + 1][i] * (i - j) + K); 
	} 
	printf("%lld
", f[N]); 
    return 0; 
}