BZOJ 1003 物流运输(最短路+DP)

容易发现DP的做法，dp[i]表示1-i天的最小代价。令cost[i][j]表示i-j天用同一条路的最小价值。

则有 dp[i]=dp[j]+cost[j+1][i]*(i-j)+k (j<i).

cost[i][j]可以用n^2次最短路求出来。由于最短路用的是堆优化dijkstra。

所以复杂度为O(n^2+m^2*ElogE).

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 1000000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1000;
//Code begin...
 
struct Edge{int p, next, w;}edge[N*N*4];
int head[N], cnt=1, cost[105][105], node[N], dis[N];
LL dp[N];
bool vis[N];
vector<PII>tim[N];
struct qnode{
    int v, c;
    qnode(int _v=0, int _c=0):v(_v),c(_c){}
    bool operator <(const qnode &r)const{return c>r.c;}
};
void add_edge(int u, int v, int w)
{
    edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;
}
void dij(int n, int start)
{
    mem(vis,0);
    FOR(i,1,n) dis[i]=INF;
    priority_queue<qnode>que;
    while (!que.empty()) que.pop();
    dis[start]=0; que.push(qnode(start,0));
    qnode tmp;
    while (!que.empty()) {
        tmp=que.top(); que.pop();
        int u=tmp.v;
        if (vis[u]) continue;
        vis[u]=true;
        for (int i=head[u]; i; i=edge[i].next) {
            int v=edge[i].p;
            if (!vis[v]&&!node[v]&&dis[v]>dis[u]+edge[i].w) {
                dis[v]=dis[u]+edge[i].w;
                que.push(qnode(v,dis[v]));
            }
        }
    }
}
int main ()
{
    int n, m, k, d, e, u, v, w;
    scanf("%d%d%d%d",&n,&m,&k,&e);
    while (e--) scanf("%d%d%d",&u,&v,&w), add_edge(u,v,w), add_edge(v,u,w);
    scanf("%d",&d);
    FOR(i,1,d) scanf("%d%d%d",&u,&v,&w), tim[u].pb(mp(v,w));
    FOR(i,1,n) FOR(j,i,n) {
        mem(node,0);
        FOR(k,1,m) FO(l,0,tim[k].size()) {
            int fi=tim[k][l].first, se=tim[k][l].second;
            if ((fi>=i&&se<=j)||(i>=fi&&i<=se)||(j>=fi&&j<=se)) {node[k]=1; break;}
        }
        dij(m,1);
        cost[i][j]=dis[m];
    }
    dp[1]=cost[1][1];
    FOR(i,2,n) {
        dp[i]=(LL)cost[1][i]*i;
        FO(j,1,i) dp[i]=min(dp[i],dp[j]+(LL)cost[j+1][i]*(i-j)+k);
    }
    printf("%lld\n",dp[n]);
    return 0;
}