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UVa--10048 Audiophobia(floyd)

2016-03-29 17:32 465 查看
UVa 10048

题意

求图中两点最大噪声值最小的路径。

题解

Floyd算法变形一下:

d[i][j]=min(d[i][j],max(d[i][k],d[k][j]))

#include <bits/stdc++.h>
using namespace std;

const int maxn = 105;
const int inf = 1000001;
int   d[maxn][maxn];

int main()
{
#ifdef LOCAL
fstream cin("data.in");
#endif // LOCAL

int C, S, Q;
int kcase = 0;

ios::sync_with_stdio(0);
cin.tie(0);
while(cin >> C >> S >> Q && C)
{
int a, b, w;
for(int i = 1; i <= C; ++i) fill(*(d + i), *(d + i) + maxn, inf);
for(int i = 1; i <= C; ++i) d[i][i] = 0;
for(int i = 1; i <= S; ++i)
{
cin >> a >> b >> w;
d[a][b] = d[b][a] = w;
}
for(int k = 1; k <= C; ++k)
for(int i = 1; i <= C; ++i)
for(int j = 1; j <= C; ++j)
d[i][j] = min(d[i][j], max(d[i][k], d[k][j]));
if(kcase++) cout << endl;
cout << "Case #" << kcase << endl;
while(Q--){
cin >> a >> b;
if(d[a][b] == inf) cout << "no path" << endl;
else cout << d[a][b] << endl;
}
}
return 0;
}
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标签:  UVa floyd easy++