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Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2445 Accepted Submission(s): 720
要拆除的边必为割边(桥)。
找到所有的割边,对每条割边e(u, v),算出e所连接的两个连通分量中的总人数之差,更新答案。
注意:题目未指明图中是否有重边,应当按照有重边处理。
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#include<bits/stdc++.h>
using namespace std;
const int N=1e4+5, M=2e4+5;
struct edge{
int to, nt;
}E[M<<1];
int head[N], sz[N], dfn[N], low[N], id, tot, ans;
void dfs(int u, int f){
dfn[u]=low[u]=++id;
for(int i=head[u]; ~i; i=E[i].nt){
int &v=E[i].to;
if(!dfn[v]){
dfs(v, i>>1);
low[u]=min(low[v], low[u]);
sz[u]+=sz[v];
if(low[v]>dfn[u]){
ans=min(ans,abs(tot-2*sz[v]));
}
}
else if(v!=i>>1&&dfn[v]<dfn[u]){
low[u]=min(low[u], dfn[v]);
}
}
}
int main(){
//freopen("in", "r", stdin);
int n, m, u, v;
for(;~scanf("%d%d", &n, &m);){
ans=INT_MAX, tot=0;
memset(head, -1, sizeof(head)), id=0;
for(int i=0; i<n; i++) scanf("%d", sz+i), tot+=sz[i];
for(;m--;){
scanf("%d%d", &u, &v);
E[id]={v, head[u]}, head[u]=id++;
E[id]={u, head[v]}, head[v]=id++;
}
memset(dfn, 0, sizeof(dfn)), id=0;
dfs(0, -1);
ans==INT_MAX?puts("impossible"):printf("%d\n", ans);
}
}
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原文地址:http://www.cnblogs.com/Patt/p/4726969.html