题目:题目链接
题意:其实就是给n个点,m条边,每条边有个权值,然后求之中的一个点,到每个点的距离的最短的消耗
分析:暴力的Floyd,直接把每一个点和每一个点相连,最后对于每一个点求最长的消耗。然后总体再求最小的消耗
代码:
#include <iostream> #include <cstdio> #include <string> #include <string.h> #include <map> #include <vector> #include <cstdlib> #include <algorithm> #include <cmath> #include <queue> #include <set> #include <stack> #include <functional> #include <fstream> #include <sstream> #include <iomanip> #include <numeric> #include <cassert> #include <bitset> #include <stack> #include <ctime> #include <list> #define INF 6666 #define max3(a,b,c) (max(a,b)>c?max(a,b):c) #define min3(a,b,c) (min(a,b)<c?min(a,b):c) #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; #define maxn 110 int QuickMod(int a,int b,int n) { int r = 1; while(b) { if(b&1) r = (r*a)%n; a = (a*a)%n; b >>= 1; } return r; } int n; int dis[maxn][maxn]; void FLOYD() { for(int k = 1; k <= n; ++k) { for(int i = 1; i <= n; ++i) { for(int j = 1; j <= n; ++j) { if(i != j && dis[i][j] > dis[i][k] + dis[k][j]) dis[i][j] = dis[i][k] + dis[k][j]; } } } int MAX; int MIN = INF; int ans; for(int i = 1; i <= n; ++i) { MAX = 0; for(int j = 1; j <= n; ++j) { if(i != j && MAX < dis[i][j]) MAX = dis[i][j]; } if(MIN > MAX) { MIN = MAX; ans = i; } } if(MIN < INF) printf("%d %d\n", ans, MIN); else printf("disjoint\n"); } int main() { while(scanf("%d", &n) && n) { mem(dis, INF); for(int i = 1; i <= n; ++i) { int m; scanf("%d", &m); for(int j = 0; j < m; ++j) { int mubiao, time; scanf("%d%d", &mubiao, &time); dis[i][mubiao] = time; } } FLOYD(); } return 0; }
顺便说一下,memset的置0和置-1可以,其他的不行的