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hdu6832（2020hdu多校6t6）
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題解
所有最短路肯定都出現(xiàn)在最小生成樹上。因?yàn)榭唆斔箍柊错樞蜻B邊，前面已經(jīng)連上的肯定最優(yōu)，所有加起來比當(dāng)前的還小。
這個(gè)我感覺求單源多匯最短路后的剩余圖其實(shí)和最小生成樹其實(shí)是一樣的。因?yàn)槿绻钚∩蓸涞慕Y(jié)論是對(duì)的，那么源點(diǎn)到所有點(diǎn)的最短路一定都在樹上，剩余圖就是這個(gè)最小生成樹。我試了試都是可以AC的。
開始初始化錯(cuò)了，少初始化了一堆東西，dfs停不下來，MLE。后來wa，因?yàn)槊杜e邊端點(diǎn)dp計(jì)數(shù)沒想清。
AC代碼
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std
;
const int modd
=1e9+7;
const int oo
=1e9+10;
const int NN
=100100;
const int MM
=200100;
int n
,m
;
int vis
[NN
],dis
[NN
];
struct edge
{int to
,cost
,from
;
}e
[MM
*2];
vector
<int> con
[NN
];
int a
[NN
];
struct point
{int id
;long long d
;
};
struct point_cmp
{bool
operator() (const point
&x
,const point
&y
)const{if(x
.d
>y
.d
)return 1;if(x
.d
==y
.d
&& x
.id
>y
.id
)return 1;return 0;}
};
priority_queue
<point
,vector
<point
> , point_cmp
> q
;
int pre
[NN
];
void dij(int origin
){for(int i
=0;i
<=n
+1;i
++)dis
[i
]=oo
;dis
[origin
]=0;point st
;st
.id
=origin
;st
.d
=0;q
.push(st
);while(!q
.empty()){point root
=q
.top();q
.pop();if(dis
[root
.id
]!=root
.d
)continue;for(int i
=0;i
<con
[root
.id
].size();i
++){edge ed
=e
[con
[root
.id
][i
]];int nex
=ed
.to
;if(dis
[nex
]>max(dis
[root
.id
],ed
.cost
) ){pre
[nex
]=con
[root
.id
][i
];dis
[nex
]=max(dis
[root
.id
],ed
.cost
);point neww
;neww
.id
=nex
;neww
.d
=dis
[nex
];q
.push(neww
);}}}
}
long long siz
[NN
],sizfa
[NN
],numson1
[NN
],numfa1
[NN
],num1
;
int f
[MM
];
void dfs(int cur
,int fa
){int up
=con
[cur
].size();numson1
[cur
]=(a
[cur
]==1);siz
[cur
]=1;for(int i
=0;i
<up
;i
++){edge ed
=e
[con
[cur
][i
]];int nex
=ed
.to
;if(f
[con
[cur
][i
]]&&nex
!=fa
){dfs(nex
,cur
);siz
[cur
]+=siz
[nex
];numson1
[cur
]+=numson1
[nex
];}}
}
long long ans
=0;
long long quan
[MM
];
void dfsfa(int cur
,int fa
,int co
){int up
=con
[cur
].size();numfa1
[cur
]=num1
-numson1
[cur
];sizfa
[cur
]=n
-siz
[cur
];if(cur
!=1){ans
+=quan
[co
]*(1ll*numfa1
[cur
]*(siz
[cur
]-numson1
[cur
])%modd
+1ll*(sizfa
[cur
]-numfa1
[cur
])*numson1
[cur
]%modd
)%modd
;ans
%=modd
;}for(int i
=0;i
<up
;i
++){edge ed
=e
[con
[cur
][i
]];int nex
=ed
.to
;if(f
[con
[cur
][i
]]&&nex
!=fa
){dfsfa(nex
,cur
,ed
.cost
);}}
}
int fa
[NN
];
int getf(int x
){if(fa
[x
]==x
)return x
;return (fa
[x
]=getf(fa
[x
]));
}
int main(){quan
[0]=1;for(int i
=1;i
<=200000;i
++){quan
[i
]=(quan
[i
-1]<<1)%modd
;}int t
;scanf("%d",&t
);while(t
--){scanf("%d%d",&n
,&m
);if(n
==0)break;for(int i
=1;i
<=n
;i
++)con
[i
].clear(),fa
[i
]=i
;for(int i
=0;i
<=m
*2;i
++)f
[i
]=0;num1
=0;for(int i
=1;i
<=n
;i
++){scanf("%d",a
+i
);num1
+=(a
[i
]==1);}int nume
=-1;for(int i
=1;i
<=m
;i
++){int x
,y
;scanf("%d%d",&x
,&y
);e
[++nume
].to
=y
;e
[nume
].cost
=i
;e
[nume
].from
=x
;con
[x
].push_back(nume
);e
[++nume
].to
=x
;e
[nume
].cost
=i
;e
[nume
].from
=y
;con
[y
].push_back(nume
);}dij(1);for(int i
=1;i
<=n
;i
++){f
[pre
[i
]]=1;f
[pre
[i
]^1]=1;}dfs(1,-1);ans
=0;dfsfa(1,-1,0);printf("%lld\n",ans
);}return 0;
}
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std
;
const int modd
=1e9+7;
const int oo
=1e9+10;
const int NN
=100100;
const int MM
=200100;
int n
,m
;
struct edge
{int to
,cost
,from
;
}e
[MM
*2];
vector
<int> con
[NN
];
int a
[NN
];
long long siz
[NN
],sizfa
[NN
],numson1
[NN
],numfa1
[NN
],num1
;
int f
[MM
];
void dfs(int cur
,int fa
){int up
=con
[cur
].size();numson1
[cur
]=(a
[cur
]==1);siz
[cur
]=1;for(int i
=0;i
<up
;i
++){edge ed
=e
[con
[cur
][i
]];int nex
=ed
.to
;if(f
[con
[cur
][i
]]&&nex
!=fa
){dfs(nex
,cur
);siz
[cur
]+=siz
[nex
];numson1
[cur
]+=numson1
[nex
];}}
}
long long ans
=0;
long long quan
[MM
];
void dfsfa(int cur
,int fa
,int co
){int up
=con
[cur
].size();numfa1
[cur
]=num1
-numson1
[cur
];sizfa
[cur
]=n
-siz
[cur
];if(cur
!=1){ans
+=quan
[co
]*(1ll*numfa1
[cur
]*(siz
[cur
]-numson1
[cur
])%modd
+1ll*(sizfa
[cur
]-numfa1
[cur
])*numson1
[cur
]%modd
)%modd
;ans
%=modd
;}for(int i
=0;i
<up
;i
++){edge ed
=e
[con
[cur
][i
]];int nex
=ed
.to
;if(f
[con
[cur
][i
]]&&nex
!=fa
){dfsfa(nex
,cur
,ed
.cost
);}}
}
int fa
[NN
];
int getf(int x
){if(fa
[x
]==x
)return x
;return (fa
[x
]=getf(fa
[x
]));
}
int main(){quan
[0]=1;for(int i
=1;i
<=200000;i
++){quan
[i
]=(quan
[i
-1]<<1)%modd
;}int t
;scanf("%d",&t
);while(t
--){scanf("%d%d",&n
,&m
);if(n
==0)break;for(int i
=1;i
<=n
;i
++)con
[i
].clear(),fa
[i
]=i
;for(int i
=0;i
<=m
*2;i
++)f
[i
]=0;num1
=0;for(int i
=1;i
<=n
;i
++){scanf("%d",a
+i
);num1
+=(a
[i
]==1);}int nume
=-1;for(int i
=1;i
<=m
;i
++){int x
,y
;scanf("%d%d",&x
,&y
);e
[++nume
].to
=y
;e
[nume
].cost
=i
;e
[nume
].from
=x
;con
[x
].push_back(nume
);e
[++nume
].to
=x
;e
[nume
].cost
=i
;e
[nume
].from
=y
;con
[y
].push_back(nume
);}for(int i
=0;i
<=nume
;i
+=2){int fx
=getf(e
[i
].from
);int fy
=getf(e
[i
].to
);if(fx
!=fy
){fa
[fx
]=fy
;f
[i
]=1;f
[i
^1]=1;}}dfs(1,-1);ans
=0;dfsfa(1,-1,0);printf("%lld\n",ans
);}return 0;
}
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