https://pintia.cn/problem-sets/994805342720868352/problems/994805358663417856
很傳統的最短路，不過要跑兩次，其實分開來的話每一問都是比較常規的題目。
#include<bits/stdc++.h>
using namespace std
;
const int N
=1e3+10;
int L
[N
][N
],T
[N
][N
],vis
[N
],dist1
[N
],dist2
[N
];
int n
,m
,st
,ed
;
vector
<int>ans1
,ans2
,path
;
int max_time
=1e9,min_cnt
=1e9;
void Dijkstra(int g
[N
][N
],int dist
[N
])
{for(int i
=0;i
<n
;i
++) dist
[i
]=0x3f3f3f3f;dist
[st
]=0;memset(vis
,0,sizeof vis
);for(int i
=0;i
<n
;i
++){int t
=-1;for(int j
=0;j
<n
;j
++)if(!vis
[j
]&&(t
==-1 || dist
[j
]<dist
[t
])) t
=j
;vis
[t
]=1;for(int j
=0;j
<n
;j
++) dist
[j
]=min(dist
[j
],dist
[t
]+g
[t
][j
]);}
}
void dfs1(int u
,int fa
,int sum
)
{if(u
==ed
){if(sum
<max_time
) max_time
=sum
,ans1
=path
;return;}for(int i
=0;i
<n
;i
++){if(i
==fa
) continue;if(dist1
[i
]==dist1
[u
]+L
[u
][i
]){path
.push_back(u
);dfs1(i
,u
,sum
+T
[u
][i
]);path
.pop_back();}}
}
void dfs2(int u
,int fa
,int sum
)
{if(u
==ed
){if(sum
<min_cnt
) ans2
=path
,min_cnt
=sum
;return;}for(int i
=0;i
<n
;i
++){if(i
==fa
) continue;if(dist2
[i
]==dist2
[u
]+T
[u
][i
]){path
.push_back(u
);dfs2(i
,u
,sum
+1);path
.pop_back();}}
}
int main(void)
{memset(L
,0x3f,sizeof L
);memset(T
,0x3f,sizeof T
);cin
>>n
>>m
; for(int i
=0;i
<m
;i
++){int a
,b
,op
,l
,t
; cin
>>a
>>b
>>op
>>l
>>t
;if(op
){L
[a
][b
]=min(L
[a
][b
],l
);T
[a
][b
]=min(T
[a
][b
],t
);}else{L
[a
][b
]=min(L
[a
][b
],l
);L
[b
][a
]=min(L
[b
][a
],l
);T
[a
][b
]=min(T
[a
][b
],t
);T
[b
][a
]=min(T
[b
][a
],t
);}}cin
>>st
>>ed
;Dijkstra(L
,dist1
);Dijkstra(T
,dist2
);dfs1(st
,-1,0);dfs2(st
,-1,0);if(ans1
!=ans2
){printf("Distance = %d: ",dist1
[ed
]);for(int i
=0;i
<ans1
.size();i
++) cout
<<ans1
[i
]<<" -> ";cout
<<ed
<<endl
;printf("Time = %d: ",dist2
[ed
]);for(int i
=0;i
<ans2
.size();i
++) cout
<<ans2
[i
]<<" -> ";cout
<<ed
<<endl
;}else{printf("Distance = %d; Time = %d: ",dist1
[ed
],dist2
[ed
]);for(int i
=0;i
<ans1
.size();i
++) cout
<<ans1
[i
]<<" -> ";cout
<<ed
<<endl
;}return 0;
}
總結
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