hdu3665 Seaside (floyd算法)

题意：　有n个镇，分别编号为0~n-1，有些镇靠海，xiaoY现在0镇，求出从0镇到海边的最短路径长度。

思路：　floyd算法。

学习点：floyd算法 

资料：

1、

http://v.youku.com/v_showMini/id_XMTA1MTg5OTQ0_ft_344.html；

 

2、

 

http://zh.wikipedia.org/wiki/%E5%BC%97%E6%B4%9B%E4%BC%8A%E5%BE%B7%E7%AE%97%E6%B3%95（维基百科上的解释很清楚）

源代码：

 #include <stdio.h> #define MAXN 11 #define INF 32767 int nMap[MAXN][MAXN]; int main() { int i, j, k; int N, nTo, nFrom; int nNum, nS, isNear; while (1 == scanf("%d", &N)) { for (i = 0; i <= N; ++i) { for (j = 0; j <= N; ++j) { nMap[i][j] = INF; } } for (nFrom = 0; nFrom != N; ++nFrom) { scanf("%d %d", &nNum, &isNear); if (isNear) nMap[nFrom][N] = 0; for (i = 0; i != nNum; ++i) { scanf("%d %d", &nTo, &nS); nMap[nFrom][nTo] = nS; } } for (i = 0; i <= N; ++i) { for (j = 0; j <= N; ++j) { for (k = 0; k <= N; ++k) { if (nMap[i][j] > nMap[i][k] + nMap[k][j]) { nMap[i][j] = nMap[i][k] + nMap[k][j]; } } } } printf("%d/n", nMap[0][N]); } return 0; }