[ZJOI2006]物流运输

2020-03-02
ZJOI

题意

共$n$天运输,每天可能封掉一些点,每次更改路线要$K$的代价,单位长度的运输要1的代价

求最小代价,$n\leq 100,|E|\leq 20$

题解

对于天数进行Dp,转移的时候求出第i+1~j天都可行的最短路即可

调试记录

边的数组开小了(maxn<<1)

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#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
const int INF = 0x3f3f3f3f;
const int maxn = 105;
using namespace std;
struct E{
int to, nxt, l;
}e[maxn * maxn << 1];
int head[maxn], tot = 0;
void addedge(int u, int v, int l){
e[++tot].to = v, e[tot].nxt = head[u], e[tot].l = l;
head[u] = tot;
}
priority_queue <pair<int, int> > q;
int dis[maxn]; bool vis[maxn];
int D, K, Q, n, m;
void Dijkstra(){
q.push(make_pair(0, 1));
for (int i = 2; i <= n; i++) dis[i] = INF; dis[1] = 0;
while (!q.empty()){
int cur = q.top().second; q.pop();
if (vis[cur]) continue; vis[cur] = 1;
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (dis[v] > dis[cur] + e[i].l){
dis[v] = dis[cur] + e[i].l;
q.push(make_pair(-dis[v], v));
}
}
}
}
int f[maxn][maxn];
int dp[maxn];
int main(){
scanf("%d%d%d%d", &D, &n, &K, &m);
for (int u, v, l, i = 1; i <= m; i++){
scanf("%d%d%d", &u, &v, &l);
addedge(u, v, l); addedge(v, u, l);
} scanf("%d", &Q);
for (int p, l, r, i = 1; i <= Q; i++){
scanf("%d%d%d", &p, &l, &r);
for (int j = l; j <= r; j++) f[p][j] = 1;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= D; j++)
f[i][j] += f[i][j - 1];
memset(dp, 0x3f, sizeof dp); dp[0] = -K;
for (int i = 1; i <= D; i++){
for (int j = 0; j < i; j++){
for (int k = 1; k <= n; k++)
vis[k] = f[k][i] - f[k][j];
Dijkstra();
if (dis[n] == INF) continue;
dp[i] = min(dp[i], dp[j] + K + dis[n] * (i - j));
}
} printf("%d\n", dp[D]);
return 0;
}