BZOJ 1003 物流运输trans dijstra+dp

题目链接:

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

题意:

题解:

首先我们必须机智的知道f[i]=min(f[i],f[j]+cost(j+1,i)+k)这个dp方程
cost(i,j)表示从第i天到第j天的最小花费
dijstra跑一发

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 9999999;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e2+10;

int n,m,k,e;
int flag[maxn][maxn],f[maxn],d[maxn],dp[maxn];
vector<pair<int,int> > G[maxn];

int cost(int s,int e){
    MS(f); 
    for(int i=1;i<=m;i++)
        d[i]=INF;

    for(int i=1; i<=m; i++){
        for(int j=s; j<=e; j++)
            if(flag[i][j])
                f[i] = 1;
    }

    queue<int> q;
    d[1] = 0;
    q.push(1);
    while(!q.empty()){
        int u = q.front(); q.pop();
        for(int i=0; i<(int)G[u].size(); i++){
            pair<int,int> p = G[u][i];
            int v = p.first, w = p.second;
            if(f[v]) continue;
            if(d[v] > d[u]+w){
                d[v] = d[u]+w;
                q.push(v);
            }
        }
    }
    // cout << d[m] << "  pp
";
    int ans = d[m]*(e-s+1);
    return ans;
}

int main(){
    scanf("%d%d%d%d",&n,&m,&k,&e);
    for(int i=0; i<e; i++){
        int u,v,w; scanf("%d%d%d",&u,&v,&w);
        G[u].push_back(MP(v,w));
        G[v].push_back(MP(u,w));
    }

    int d = read();
    for(int i=0; i<d; i++){
        int u,ii,jj;
        scanf("%d%d%d",&u,&ii,&jj);
        for(int j=ii; j<=jj; j++)
            flag[u][j] = 1;
    }
    // cout << "hh
";
    for(int i=1; i<=n; i++){
        dp[i] = cost(1,i);
        // cout << dp[i] << "  kk
";
        for(int j=1; j<i; j++)
            dp[i] = min(dp[i],dp[j]+cost(j+1,i)+k);
    }

    cout << dp[n] << endl;

    return 0;
}