poj 1062

用dfs找最短路。
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 #include<cmath>
 5 using namespace std;
 6 const int inf=0x3f3f3f3f;
 7 int mp[120][120],dif[120],mi[120],mx[120];
 8 int m,n,out;
 9 int my_min(int x,int y){ return x<y?x:y; }
10 bool cheak(int x){ return x>m?true:false; }
11 bool cheak2(int x,int y){
12     if(dif[x]<=mi[y])
13         if((mi[y]-dif[x]<=m)&&(mx[y]-dif[x]<=m))
14             return true;
15     if(dif[x]>=mx[y])
16         if((dif[x]-mx[y]<=m)&&(dif[x]-mi[y]<=m))
17             return true;
18     if(dif[x]<=mx[y]&&dif[x]>=mi[y])
19         return true;
20     return false;
21 }
22 void dfs(int ans, int next)
23 {
24     if(ans>=out)
25         return ;
26     for(int i=1;i<=n;i++){
27         if(mp[next][i]!=inf&&next==i){
28             if(out>ans+mp[next][i]&&cheak2(i,next))
29                 out=ans+mp[next][i];
30         }
31         if(mp[next][i]!=inf&&next!=i){
32             if(cheak2(i,next)){
33                 int temp=mi[i],temp2=mx[i];
34                 if(dif[i]<=mi[next]){
35                     mi[i]=dif[i];
36                     mx[i]=mx[next];
37                 }
38                 else if(dif[i]>=mx[next]){
39                     mx[i]=dif[i];
40                     mi[i]=mi[next];
41                 }
42                 else if(dif[i]<=mx[next]&&dif[i]>=mi[next]){
43                     mx[i]=mx[next];
44                     mi[i]=mi[next];
45                 }
46                 dfs(ans+mp[next][i],i);
47                 mi[i]=temp;
48                 mx[i]=temp2;
49             }
50         }
51     }
52 }
53 
54 int main()
55 {
56     scanf("%d%d",&m,&n);
57     memset( mp, inf, sizeof mp);
58     for(int i=1;i<=n;i++){
59         int p,l,x;
60         scanf("%d%d%d",&p,&l,&x);
61         mp[i][i]=p;
62         dif[i]=mi[i]=mx[i]=l;
63         bool flag=cheak(sqrt(dif[i]-dif[1]));
64         for(int j=1;j<=x;j++){
65             int t,v;
66             scanf("%d%d",&t,&v);
67             if(flag)
68                 continue;
69             mp[i][t]=my_min(mp[i][t],v);
70         }
71     }
72     out=mp[1][1];
73     dfs(0,1);
74     printf("%d\n",out);
75     return 0;
76 }
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