ZOJ 2797 ||POJ 2472 floyd变形

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记住一定要初始化dis[][]，因为这边wa了几次。

题意：求起点1到终点n不被抓的概率。也就是要是输出的结果最大。只要把floyd状态方程改为*就行了。如下所示：

dis[i][j]=max(dis[i][j],dis[i][k]*dis[k][j]).

切记初始化dis[i][j]=0;

要求最大则初始化为-INF 或者0.不能是INF

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,m; 66 const int maxn = 1000+10; 67 double dis[maxn][maxn]; 68 int s,e; 69 double w; 70 void floyd(){ 71     for(int k=1;k<=n;k++){ 72         for(int i=1;i<=n;i++){ 73             for(int j=1;j<=n;j++){ 74                 dis[i][j]=max(dis[i][j],dis[i][k]*dis[k][j]); 75             } 76         } 77     } 78 } 79  80 int main(){ 81    // int n,m; 82     while(scanf("%d",&n)!=EOF){ 83         if(n==0)  break; 84         scanf("%d",&m); 85         for(int i=1;i<=n;i++){ 86             for(int j=1;j<=n;j++){ 87                 dis[i][j]=0; 88             } 89             dis[i][i]=0; 90         } 91         for(int i=1;i<=m;i++){ 92             scanf("%d%d%lf",&s,&e,&w); 93             dis[s][e]=dis[e][s]=(double)(w/100); 94             //printf("****%lf\n",dis[s][e]); 95         } 96         floyd(); 97  98         printf("%.6lf percent\n",dis[1][n]*100); 99     }100     return 0;101 }