【Beijing WC2012】 冻结

【题目链接】

点击打开链接

【算法】

dist[i][j]表示到达i号城市，使用了j次魔法，所用时间的最小值

那么，dist[i][j]可以转移到dist[k][j+1]和dist[k][j],一边spfa一边dp，即可

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 55
#define MAXM 2010

const int INF = 1e9;

int i,n,m,k,a,b,t,ans = INF,tot;
int head[MAXN],inq[MAXN][MAXN],dist[MAXN][MAXN];

struct Edge
{
        int to,cost,nxt;
} e[MAXM];

template <typename T> inline void read(T &x)
{
    int f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < )
    {
        putchar('-');
        x = -x;
    }
    if (x > ) write(x/);
    putchar(x%+'');
}
template <typename T> inline void writeln(T x)
{
    write(x);
    puts("");
}
inline void add(int u,int v,int w)
{
        tot++;
        e[tot] = (Edge){v,w,head[u]};
        head[u] = tot;
}
inline void spfa(int s)
{
        int i,j;
        queue< pair<int,int> > q;
        pair<int,int> cur;
        for (i = ; i <= n; i++)
        {
                for (j = ; j <= k; j++)
                {
                        dist[i][j] = INF;
                }
        }
        dist[][] = ;
        q.push(make_pair(,));
        inq[][] = ;
        while (!q.empty())
        {
                cur = q.front();
                inq[cur.first][cur.second] = false;
                q.pop();
                for (i = head[cur.first]; i; i = e[i].nxt)
                {
                        if (dist[cur.first][cur.second] + e[i].cost < dist[e[i].to][cur.second])
                        {
                                dist[e[i].to][cur.second] = dist[cur.first][cur.second] + e[i].cost;
                                if (!inq[e[i].to][cur.second])
                                {
                                        inq[e[i].to][cur.second] = ;
                                        q.push(make_pair(e[i].to,cur.second));
                                }
                        }
                        if ((cur.second +  <= k) && (dist[cur.first][cur.second] + e[i].cost /  < dist[e[i].to][cur.second+]))
                        {
                                dist[e[i].to][cur.second+] = dist[cur.first][cur.second] + e[i].cost / ;
                                if (!inq[e[i].to][cur.second+])
                                {
                                        inq[e[i].to][cur.second+] = ;
                                        q.push(make_pair(e[i].to,cur.second+));
                                }
                        }
                }
        }
}

int main() {

        read(n); read(m); read(k);
        for (i = ; i <= m; i++)
        {
                read(a); read(b); read(t);
                add(a,b,t);
                add(b,a,t);
        }
        spfa();

        for (i = ; i <= k; i++) ans = min(ans,dist[n][i]);

        writeln(ans);

        return ;

}