# BZOJ1003: [ZJOI2006]物流运输(最短路+DP)

## 题目：

1003: [ZJOI2006]物流运输

## 解析：

$cost[i][j]$表示第$i$天到第$j$天的花费

$f[i] = min(f[i], f[j]+k+cost[j+1][i]*(i-j))$

## 代码：

#include <bits/stdc++.h>

using namespace std;

const int N = 1010;
const int INF = 0x3f3f3f3f;

int n, m, k, E, num, t;
int head[N], dis[N];

long long f[N], cost[N][N];

bool vis[N], no[N][N], mark[N];

struct node {
int v, nx, w;
} e[N];

struct edge {
int id, dis;
bool operator <(const edge &oth) const {
return dis > oth.dis;
}
};

inline void add(int u, int v, int w) {
e[++num] = (node) {v, head[u], w}, head[u] = num;
}

priority_queue<edge>q;
void dijkstra(int s, int from, int to) {
memset(dis, INF, sizeof dis);
memset(vis, 0, sizeof vis);
memset(mark, 0, sizeof mark);
for (int i = 1; i <= m; ++i)
for (int j = from; j <= to; ++j)
if (no[i][j]) mark[i] = 1;
dis[s] = 0;
q.push((edge) {s, 0});
while (!q.empty()) {
int u = q.top().id;
q.pop();
if (vis[u]) continue;
vis[u] = 1;
for (int i = head[u]; ~i; i = e[i].nx) {
int v = e[i].v;
if (mark[v]) continue;
if (dis[v] > dis[u] + e[i].w)
q.push((edge) {v, dis[v] = dis[u] + e[i].w});
}
}
}

signed main() {
memset(head, -1, sizeof head);
cin >> n >> m >> k >> E;
for (int i = 1, x, y, z; i <= E; ++i) {
cin >> x >> y >> z;
add(x, y, z), add(y, x, z);
}
cin >> t;
for (int i = 1, x, y, z; i <= t; ++i) {
cin >> x >> y >> z;
for (int j = y; j <= z; ++j) no[x][j] = 1;
}
for (int i = 1; i <= n; ++i)
for (int j = i; j <= n; ++j) {
dijkstra(1, i, j);
cost[i][j] = dis[m];
}
for (int i = 1; i <= n; ++i) ;
for (int i = 1; i <= n; ++i) {
f[i] = cost[1][i] * i;
for (int j = 1; j <= i; ++j) {
f[i] = min(f[i], f[j] + k + cost[j + 1][i] * (i - j));
}
}
cout << f[n];
}
posted @ 2019-08-13 17:16 Chrety 阅读(...) 评论(...) 编辑 收藏