【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 = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x/10);
    putchar(x%10+'0');
}
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 = 1; i <= n; i++)
        {
                for (j = 0; j <= k; j++)
                {
                        dist[i][j] = INF;
                }
        }
        dist[1][0] = 0;
        q.push(make_pair(1,0));
        inq[1][0] = 1;
        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] = 1;
                                        q.push(make_pair(e[i].to,cur.second)); 
                                } 
                        }
                        if ((cur.second + 1 <= k) && (dist[cur.first][cur.second] + e[i].cost / 2 < dist[e[i].to][cur.second+1]))
                        {
                                dist[e[i].to][cur.second+1] = dist[cur.first][cur.second] + e[i].cost / 2;
                                if (!inq[e[i].to][cur.second+1])
                                {
                                        inq[e[i].to][cur.second+1] = 1;
                                        q.push(make_pair(e[i].to,cur.second+1));
                                }
                        }
                }
        }        
}

int main() {
        
        read(n); read(m); read(k);
        for (i = 1; i <= m; i++)
        {
                read(a); read(b); read(t);
                add(a,b,t);
                add(b,a,t);
        }
        spfa(1);

        for (i = 0; i <= k; i++) ans = min(ans,dist[n][i]);
        
        writeln(ans);
                
        return 0;
    
}