卧薪尝胆,厚积薄发。
SHOI2010 最小生成树
Date: Sun Mar 24 16:16:28 CST 2019 In Category: NoCategory

Description:

给一个带权无向连通图,每次可以给一条边边权 $+1$ ,问至少操作多少次可以使得给出的某条边一定在 $MST$ 上。
$1\leqslant n\leqslant 500,1\leqslant m\leqslant 800$

Solution:

在原图中把这条边删掉,其他边的权值是这条边的权值 $+1-$ 原来的权值,然后分别从源 $/$ 汇朝这条边的两个端点连边跑最小割就是答案。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cctype>
#include<cstring>
using namespace std;
inline int rd()
{
register int res = 0,f = 1;register char c = getchar();
while(!isdigit(c)){if(c == '-')f = -1;c = getchar();}
while(isdigit(c))res = (res << 1) + (res << 3) + c - '0',c = getchar();
return res * f;
}
int n,m,id;
#define MAXN 510
#define MAXM 810
struct edges
{
int u,v,w;
}es[MAXM];
#define MAXP MAXN
#define MAXE (MAXM * 2)
struct edge
{
int to,nxt;
int f;
}e[MAXE << 1];
int edgenum = 0;
int lin[MAXP] = {0};
void add(int a,int b,int f)
{//cout << a << " " << b << " " << f << endl;
e[edgenum] = (edge){b,lin[a],f};lin[a] = edgenum++;
e[edgenum] = (edge){a,lin[b],0};lin[b] = edgenum++;
return;
}
int S,T;
#define INF 0x3f3f3f3f
int ch[MAXP];
bool BFS()
{
memset(ch,-1,sizeof(ch));ch[S] = 0;
queue<int> q;q.push(S);
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = lin[k];i != -1;i = e[i].nxt)
{
if(ch[e[i].to] == -1 && e[i].f)
{
ch[e[i].to] = ch[k] + 1;
q.push(e[i].to);
}
}
}
return (ch[T] != -1);
}
int flow(int k,int f)
{
if(k == T)return f;
int r = 0;
for(int i = lin[k];i != -1 && f > r;i = e[i].nxt)
{
if(ch[e[i].to] == ch[k] + 1 && e[i].f)
{
int l = flow(e[i].to,min(e[i].f,f - r));
e[i].f -= l;r += l;e[i ^ 1].f += l;
}
}
if(r == 0)ch[k] = -1;
return r;
}
int dinic()
{
int ans = 0,r;
while(BFS())while(r = flow(S,INF))ans += r;
return ans;
}
int main()
{
memset(lin,-1,sizeof(lin));
scanf("%d%d%d",&n,&m,&id);
int val;
for(int i = 1;i <= m;++i)
{
es[i].u = rd();es[i].v = rd();es[i].w = rd();
if(i == id)val = es[i].w;
}
S = n + 1;T = S + 1;
for(int i = 1;i <= m;++i)
{
if(i == id)
{
add(S,es[i].u,INF);
add(es[i].v,T,INF);
}
else
{
add(es[i].u,es[i].v,max(0,val + 1 - es[i].w));
add(es[i].v,es[i].u,max(0,val + 1 - es[i].w));
}
}
cout << dinic() << endl;
return 0;
}
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