UVA - 10048 Audiophobia Floyd
2017-03-03 14:20
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思路:套用Floyd算法思想,d(i, j) = min(d(i,j), max(d(i,k), d(k,j)),就能很方便求得任意两点之间的最小噪音路径。
AC代码
如有不当之处欢迎指出!
AC代码
#include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <utility> #include <string> #include <iostream> #include <map> #include <set> #include <vector> #include <queue> #include <stack> using namespace std; #pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10 #define inf 0x3f3f3f3f #define PI pair<int, int> typedef long long LL; const int maxn = 100 + 5; int d[maxn][maxn]; int main() { int C, S, Q, kase = 0; while(scanf("%d%d%d", &C, &S, &Q) == 3 && C+S+Q) { if(kase++) printf("\n"); memset(d, inf, sizeof(d)); int x, y, cost; for(int i = 0; i < S; ++i) { scanf("%d%d%d", &x, &y, &cost); d[x][y] = d[y][x] = cost; //无向图 } //Floyd for(int k = 1; k <= C; ++k) for(int i = 1; i <= C; ++i) for(int j = 1; j <= C; ++j) { if(d[i][k] < inf && d[k][j] < inf) { int w = max(d[i][k], d[j][k]); d[i][j] = min(d[i][j], w); } } printf("Case #%d\n", kase); for(int i = 0; i < Q; ++i) { scanf("%d%d", &x, &y); if(d[x][y] == inf) printf("no path\n"); else printf("%d\n", d[x][y]); } } return 0; }
如有不当之处欢迎指出!
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