poj 3635 Full Tank?

最近发现写程序就是策略的体现。

这题我看了一个头两个大，看了别人的解题报告，才知道是一种bfs dijkstra的结合体，给我一万个脑袋都想不出来，虽然那两种策略之前都略知一二，还是得不断做题。

#include<iostream>#include<queue>using namespace std;#define N 1005#define inf 100000000struct node{int ver,fuel,cost;};struct cmp{bool operator()(node a,node b){return a.cost>b.cost;}};bool use[N][105];int dp[N][105],p[N];struct edge{int ver,dist;edge *next;};edge *gragh[N];priority_queue<node,vector<node>,cmp>q;void add_edge(int u,int v,int d){edge *ptr=new edge;ptr->ver=v;ptr->dist=d;ptr->next=gragh[u];gragh[u]=ptr;return ;}int dijkstra(int s,int e,int cap){int cost,fuel,u,v,d;node a,temp;a.ver=s;a.fuel=a.cost=0;while(!q.empty())q.pop();q.push(a);while(!q.empty()){a=q.top();q.pop();u=a.ver,fuel=a.fuel,cost=a.cost;use[u][fuel]=1;if(u==e)return cost;if(fuel+1<=cap&&!use[u][fuel+1]&&dp[u][fuel+1]>dp[u][fuel]+p[u]){dp[u][fuel+1]=dp[u][fuel]+p[u];temp.ver=u,temp.fuel=fuel+1,temp.cost=dp[u][fuel+1];q.push(temp);}for(edge *ptr=gragh[u];ptr;ptr=ptr->next){v=ptr->ver,d=ptr->dist;if(fuel>=d&&!use[v][fuel-d]&&dp[v][fuel-d]>cost){dp[v][fuel-d]=cost;temp.ver=v,temp.fuel=fuel-d,temp.cost=dp[v][fuel-d];q.push(temp);}}}return -1;}int main(){int n,m,i,j,d,q,s,e,cap,k;scanf("%d%d",&n,&m);for(k=0;k<n;k++){scanf("%d",p+k);gragh[k]=NULL;}for(k=0;k<m;k++){scanf("%d%d%d",&i,&j,&d);add_edge(i,j,d);add_edge(j,i,d);}scanf("%d",&q);while(q--){scanf("%d%d%d",∩,&s,&e);memset(use,0,sizeof(use));for(i=0;i<n;i++)for(j=0;j<=100;j++)dp[i][j]=inf;dp[s][0]=0;i=dijkstra(s,e,cap);if(i==-1)printf("impossible/n");elseprintf("%d/n",i);}return 0;}

其实这题我一开始是用深搜的，果断TLE，贴一下那个代码

#include<iostream>using namespace std;#define N 1005#define inf 1000000000bool visited[N];int p[N],best,mind[N];struct node{node *next;int dist,v;};node *gragh[N];void add_edge(int u,int v,int d){node *ptr=new node;ptr->v=v;ptr->dist=d;ptr->next=gragh[u];gragh[u]=ptr;return ;}void init(int n,int m){int i,u,v,d;node *ptr;for(i=0;i<n;i++)scanf("%d",p+i);for(i=0;i<n;i++)gragh[i]=NULL;for(i=0;i<m;i++){scanf("%d%d%d",&u,&v,&d);add_edge(u,v,d);add_edge(v,u,d);}for(i=0;i<n;i++)mind[i]=inf;for(i=0;i<n;i++)for(ptr=gragh[i];ptr;ptr=ptr->next)if(ptr->dist<mind[i])mind[i]=ptr->dist;return ;}void dfs(int cap,int remain,int mon,int v,int end){int i;node *ptr;visited[v]=1;if(mon>=best)return ;if(v==end&&mon<best){best=mon;return ;}for(i=mind[v]-remain>0?mind[v]-remain:0;i<=cap;i++)for(ptr=gragh[v];ptr;ptr=ptr->next)if(!visited[ptr->v]){if(remain+i<ptr->dist)continue;dfs(cap,remain+i-ptr->dist,mon+p[v]*i,ptr->v,end);visited[ptr->v]=0;}return ;}int main(){int n,m,q,s,t,c;scanf("%d%d",&n,&m);init(n,m);scanf("%d",&q);for(int i=0;i<q;i++){scanf("%d%d%d",&c,&s,&t);memset(visited,0,sizeof(visited));best=inf;dfs(c,0,0,s,t);if(best==inf)printf("impossible/n");elseprintf("%d/n",best);}return 0;}