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UVA-10048 Audiophobia(最短路上的最小值,Folyd)

2016-05-24 13:31 507 查看
原有的Floyd算法应该是:

for(int k = 1;k <= N;k++)
for(int p = 1;p <= N;p++)
for(int q = 1;q <= N;q++)
{
if(d[p][k] == INF || d[k][q] == INF)
continue;
d[p][q] = min(d[p][q],d[p][k] + d[k][q]));
}


此处只需要把加号改成max即可

//#include <bits/stdc++.h>
//#define _ ios_base::sync_with_stdio(0);cin.tie(0);
#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <bitset>
#include <string>
#include <numeric>
#include <algorithm>
#include <functional>
#include <iterator>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <complex>
#include <ctime>

#define INF 0x3f3f3f3f
#define eps 1e-8
typedef long long LL;
const double pi = acos(-1.0);
const long long mod = 1e9 + 7;
using namespace std;

int M,N,Q;

int d[1005][1005];

int main()
{
int T = 0;
//freopen("int.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(scanf("%d %d %d",&N,&M,&Q))
{
if(!N && !M && !Q)
break;
for(int i = 1;i <= N;i++)
for(int j = 1;j <= N;j++)
d[i][j] = (i == j) ? 0 : INF;
int l,r,n;
for(int i = 0;i < M;i++){
scanf("%d %d %d",&l,&r,&n);
d[l][r] = d[r][l] = n;
}
for(int k = 1;k <= N;k++)
for(int p = 1;p <= N;p++)
for(int q = 1;q <= N;q++)
{
if(d[p][k] == INF || d[k][q] == INF)
continue;
d[p][q] = min(d[p][q],max(d[p][k],d[k][q]));
}
if(T)
puts("");
printf("Case #%d\n",++T);
while(Q--)
{
int c1,c2;
scanf("%d %d",&c1,&c2);
if(d[c1][c2] == INF)
puts("no path");
else
printf("%d\n",d[c1][c2]);
}
}
return 0;
}
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