ZOJ 1082 floyd

题意:求从经济人散布谣言到所有人需要的最短时间的最大值

具体详解看我原来的解法http://blog.sina.com.cn/s/blog_99ca2df50101807p.html

View Code

  1 // I'm the Topcoder  2 //C  3 #include <stdio.h>  4 #include <stdlib.h>  5 #include <string.h>  6 #include <ctype.h>  7 #include <math.h>  8 #include <time.h>  9 //C++ 10 #include <iostream> 11 #include <algorithm> 12 #include <cstdio> 13 #include <cstdlib> 14 #include <cmath> 15 #include <cstring> 16 #include <cctype> 17 #include <stack> 18 #include <string> 19 #include <list> 20 #include <queue> 21 #include <map> 22 #include <vector> 23 #include <deque> 24 #include <set> 25 using namespace std; 26  27 //*************************OUTPUT************************* 28 #ifdef WIN32 29 #define INT64 "%I64d" 30 #define UINT64 "%I64u" 31 #else 32 #define INT64 "%lld" 33 #define UINT64 "%llu" 34 #endif 35  36 //**************************CONSTANT*********************** 37 #define INF 0x3f3f3f3f 38 #define eps 1e-8 39 #define PI acos(-1.) 40 #define PI2 asin (1.); 41 typedef long long LL; 42 //typedef __int64 LL;   //codeforces 43 typedef unsigned int ui; 44 typedef unsigned long long ui64; 45 #define MP make_pair 46 typedef vector<int> VI; 47 typedef pair<int, int> PII; 48 #define pb push_back 49 #define mp make_pair 50  51 //***************************SENTENCE************************ 52 #define CL(a,b) memset (a, b, sizeof (a)) 53 #define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b)) 54 #define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c)) 55  56 //****************************FUNCTION************************ 57 template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); } 58 template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; } 59 template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; } 60  61 // aply for the memory of the stack 62 //#pragma comment (linker, "/STACK:1024000000,1024000000") 63 //end 64  65 int n; 66 const int maxn = 100+10; 67 int dis[maxn][maxn]; 68 int cost[maxn]; 69 int maxx=-INF,minx=INF; 70  71 void floyd(){ 72     for(int k=1;k<=n;k++){ 73         for(int i=1;i<=n;i++){ 74             for(int j=1;j<=n;j++){ 75                 dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]); 76             } 77         } 78     } 79 } 80  81 int main(){ 82     while(scanf("%d",&n)!=EOF){ 83         if(n==0 ) break; 84         for(int i=1;i<=n;i++){ 85             for(int j=1;j<=n;j++){ 86                 dis[i][j]=INF; 87             } 88             dis[i][i]=0; 89         } 90         int m; 91         int to,w; 92         for(int i=1;i<=n;i++){ 93             scanf("%d",&m); 94             for(int j=1;j<=m;j++){ 95                 scanf("%d%d",&to,&w); 96                 dis[i][to]=w; 97             } 98         } 99         floyd();100         int pos;101         minx=INF;102         for(int i=1;i<=n;i++){103             maxx=-INF;104             for(int j=1;j<=n;j++){105                 if(dis[i][j]>maxx){106                     //dis[i][j]=maxx;107                     maxx=dis[i][j];108                 }109             }110             if(maxx<minx){111                 pos=i;112                 minx=maxx;113             }114         }115         if(minx==INF) printf("disjoint\n");116         else printf("%d %d\n",pos,minx);117 118     }119     return 0;120 }