UVA 10269 Adventure of Super Mario 最短路

（i,j)表示当前在结点j，还剩下i次使用魔法的机会。

以(i,j)为结点建图，求最短路。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("in.txt","r",stdin);
    //freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}
struct HeapNode
{
    int d,u;
    bool operator < (const HeapNode &ant) const
    {
        return d>ant.d;
    }
};
struct Edge
{
    int from,to;
    int dist;
};
const int maxn=1205;
struct Dijksta
{
    int n;
    vector<int> g[maxn];
    vector<Edge> edge;
    int done[maxn];
    int d[maxn];

    void init(int n)
    {
        this->n=n;
        for(int i=0;i<=n;i++)
            g[i].clear();
        edge.clear();
    }

    void add(int u,int v,int w)
    {
        Edge e=(Edge){u,v,w};
        edge.push_back(e);
        g[u].push_back(edge.size()-1);
    }

    void solve(int s)
    {
        for(int i=0;i<=n;i++)d[i]=INF;
        memset(done,0,sizeof(done));
        priority_queue<HeapNode> q;
        q.push((HeapNode){0,s});
        d[s]=0;
        while(!q.empty())
        {
            HeapNode x=q.top();q.pop();
            if(done[x.u])continue;
            int u=x.u;
            done[u]=1;
            for(int i=0;i<g[u].size();i++)
            {
                Edge &e=edge[g[u][i]];
                if(d[u]+e.dist<d[e.to])
                {
                    d[e.to]=d[u]+e.dist;
                    q.push((HeapNode){d[e.to],e.to});
                }
            }
        }
    }
};

Dijksta solver;
int g1[101][101],g2[101][101];
int A,B,L,K,M;

void floyd()
{
    for(int k=1;k<=A+B;k++)
        for(int i=1;i<=A+B;i++)
            for(int j=1;j<=A+B;j++)
                if(k<=A)
                    g1[i][j]=min(g1[i][j],g1[i][k]+g1[k][j]);
}

int id[11][111];
int vcnt;
int ID(int k,int u)
{
    int &x=id[k][u];
    if(x==0)x=++vcnt;
    return x;
}

void construct()
{
    memset(id,0,sizeof(id));
    vcnt=0;

    solver.init((A+B)*(K+1));
    for(int i=K;i>=0;i--)
        for(int u=1;u<=A+B;u++)
        {
            for(int v=1;v<=A+B;v++)if(g2[u][v]!=INF)
                solver.add(ID(i,u),ID(i,v),g2[u][v]);
            if(i!=0)for(int v=1;v<=A+B;v++)if(g1[u][v]<=L)
                solver.add(ID(i,u),ID(i-1,v),0);
        }

}
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        scanf("%d%d%d%d%d",&A,&B,&M,&L,&K);
        memset(g1,INF,sizeof(g1));
        for(int i=1;i<=M;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            g1[u][v]=g1[v][u]=w;
        }
        memcpy(g2,g1,sizeof(g1));
        floyd();
        construct();
        solver.solve(ID(K,A+B));
        int ans=INF;
        for(int i=0;i<=K;i++)
            ans=min(ans,solver.d[ID(i,1)]);
        printf("%d
",ans);
    }
    return 0;
}