卧薪尝胆,厚积薄发。
USACO2014Feb GOLD Roadblock
Date: Tue Sep 04 10:02:45 CST 2018 In Category: NoCategory

Description:

把一条边的边权变成原来的二倍使得从 $1$ 到 $n$ 的最短路最长。
$1\le n \le 250,1\le m \le250000$

Solution:

更改的边权一定在原来的最短路上,而最短路最多只有 $n-1$ 条边,可以随便求出来一个最短路,暴力枚举每一条边更改边权再暴力跑最短路。
时间复杂度 $O(n\times SPFA(km$ ~ $nm))$

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
inline int read()
{
register int res = 0;
register char c = getchar();
while(!isdigit(c))c = getchar();
while(isdigit(c))
{
res = (res << 1) + (res << 3) + c - '0';
c = getchar();
}
return res;
}
int n,m;
#define MAXN 250000
struct edge
{
int to,nxt,v;
}e[MAXN << 1];
int edgenum = 0;
int lin[MAXN] = {0};
void add(int a,int b,int c)
{
e[edgenum].to = b;e[edgenum].v = c;e[edgenum].nxt = lin[a];lin[a] = edgenum;++edgenum;
e[edgenum].to = a;e[edgenum].v = c;e[edgenum].nxt = lin[b];lin[b] = edgenum;++edgenum;
return;
}
int d[MAXN],pre[MAXN];
bool v[MAXN];
int SPFA()
{
queue<int> q;
memset(d,0x3f,sizeof(d));
memset(v,false,sizeof(v));
q.push(1);d[1] = 0;
while(!q.empty())
{
int k = q.front();q.pop();
v[k] = false;
for(int i = lin[k];i != -1;i = e[i].nxt)
{
if(d[e[i].to] > d[k] + e[i].v)
{
d[e[i].to] = d[k] + e[i].v;
pre[e[i].to] = i;
if(!v[e[i].to])
{
v[e[i].to] = true;
q.push(e[i].to);
}
}
}
}
return d[n];
}
vector<int> g;
int main()
{
memset(lin,-1,sizeof(lin));
scanf("%d%d",&n,&m);
int a,b,c;
for(int i = 1;i <= m;++i)
{
a = read();b = read();c = read();
add(a,b,c);
}
int len = SPFA();
int ans = 0;
for(int i = n;i != 1;i = e[pre[i] ^ 1].to)
{
g.push_back(pre[i]);
}
for(int i = 0;i < g.size();++i)
{
int l = e[g[i]].v;
e[g[i]].v = e[g[i] ^ 1].v = 2 * l;
ans = max(ans,SPFA());
e[g[i]].v = e[g[i] ^ 1].v = l;
}
cout << ans - len << endl;
return 0;
}
In tag: 图论-SPFA
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