UVA 10269 Adventure of Super Mario 最短路-白红宇

UVA 10269 Adventure of Super Mario 最短路

阅读量：6839 次

发布时间：2019-06-26

本文共 2238 字，大约阅读时间需要 7 分钟。

（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
      
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                    using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair
                    
                      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 3.1415926535898template
                     
                       T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c));}template
                      
                        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!='\n')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
                       
                         g[maxn]; vector
                        
                          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
                         
                           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
                          
                           =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\n",ans); } return 0;}