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  • 1003: [ZJOI2006]物流运输 最短路+dp

    https://www.lydsy.com/JudgeOnline/problem.php?id=1003

    数据范围很小,怎么瞎搞都行,n方dp,然后跑出最短路暴力转移,需要注意的是不能使用的可能有多个区间

    /**************************************************************
        Problem: 1003
        User: walfy
        Language: C++
        Result: Accepted
        Time:180 ms
        Memory:1400 kb
    ****************************************************************/
     
    //#pragma comment(linker, "/stack:200000000")
    //#pragma GCC optimize("Ofast,no-stack-protector")
    //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    //#pragma GCC optimize("unroll-loops")
    #include<bits/stdc++.h>
    #define fi first
    #define se second
    #define mp make_pair
    #define pb push_back
    #define pi acos(-1.0)
    #define ll long long
    #define mod 1000000007
    #define C 0.5772156649
    #define ls l,m,rt<<1
    #define rs m+1,r,rt<<1|1
    #define pil pair<int,ll>
    #define pli pair<ll,int>
    #define pii pair<int,int>
    #define cd complex<double>
    #define ull unsigned long long
    #define base 1000000000000000000
    #define fio ios::sync_with_stdio(false);cin.tie(0)
     
    using namespace std;
     
    const double eps=1e-6;
    const int N=100+10,maxn=5000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;
     
    struct edge{
        int to,Next,c;
    }e[maxn];
    int head[N],cnt,ban[N][N];
    void add(int u,int v,int c)
    {
        e[cnt].to=v;
        e[cnt].c=c;
        e[cnt].Next=head[u];
        head[u]=cnt++;
    }
    void init()
    {
        cnt=0;
        memset(head,-1,sizeof head);
    }
    priority_queue<pair<int,int> >q;
    int dis[N],dp[N];
    int n,m,K,E;
    int dij(int be,int en)
    {
        q.push(mp(0,1));
        memset(dis,inf,sizeof dis);
        dis[1]=0;
        while(!q.empty())
        {
            pii u=q.top();
            q.pop();
            int x=u.se;
            if(dis[x]<u.fi)continue;
            for(int i=head[x];~i;i=e[i].Next)
            {
                int To=e[i].to;
                bool ok=1;
                for(int j=be;j<=en;j++)
                    if(ban[To][j])
                        ok=0;
                if(ok)
                {
                    if(dis[To]>dis[x]+e[i].c)
                    {
                        dis[To]=dis[x]+e[i].c;
                        q.push(mp(dis[To],To));
                    }
                }
            }
        }
        if(dis[m]==inf)return -1;
        else return dis[m]*(en-be+1);
    }
    int main()
    {
        scanf("%d%d%d%d",&n,&m,&K,&E);
        init();
        for(int i=0;i<E;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        int d;scanf("%d",&d);
        while(d--)
        {
            int p,a,b;scanf("%d%d%d",&p,&a,&b);
            for(int i=a;i<=b;i++)ban[p][i]=1;
        }
        memset(dp,inf,sizeof dp);
        dp[0]=-K;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<i;j++)
            {
                int dis=dij(j+1,i);
                if(dis>0)dp[i]=min(dp[i],dp[j]+dis+K);
            }
        }
        printf("%d
    ",dp[n]);
        return 0;
    }
    /********************
     
    ********************/
    View Code
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  • 原文地址:https://www.cnblogs.com/acjiumeng/p/9056827.html
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