ZOJ 2797 ||POJ 2472 floyd变形

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

// I'm the Topcoder
//C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
//C++
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <stack>
#include <string>
#include <list>
#include <queue>
#include <map>
#include <vector>
#include <deque>
#include <set>
using namespace std;

//*************************OUTPUT*************************
#ifdef WIN32
#define INT64 "%I64d"
#define UINT64 "%I64u"
#else
#define INT64 "%lld"
#define UINT64 "%llu"
#endif

//**************************CONSTANT***********************
#define INF 0x3f3f3f3f
#define eps 1e-8
#define PI acos(-1.)
#define PI2 asin (1.);
typedef long long LL;
//typedef __int64 LL;   //codeforces
typedef unsigned int ui;
typedef unsigned long long ui64;
#define MP make_pair
typedef vector<int> VI;
typedef pair<int, int> PII;
#define pb push_back
#define mp make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b, sizeof (a))
#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))

//****************************FUNCTION************************
template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }
template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }
template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }

// aply for the memory of the stack
//#pragma comment (linker, "/STACK:1024000000,1024000000")
//end

int n,m;
const int maxn = 1000+10;
double dis[maxn][maxn];
int s,e;
double w;
void floyd(){
for(int k=1;k<=n;k++){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
dis[i][j]=max(dis[i][j],dis[i][k]*dis[k][j]);
}
}
}
}

int main(){
// int n,m;
while(scanf("%d",&n)!=EOF){
if(n==0)  break;
scanf("%d",&m);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
dis[i][j]=0;
}
dis[i][i]=0;
}
for(int i=1;i<=m;i++){
scanf("%d%d%lf",&s,&e,&w);
dis[s][e]=dis[e][s]=(double)(w/100);
//printf("****%lf\n",dis[s][e]);
}
floyd();

printf("%.6lf percent\n",dis[1]
*100);
}
return 0;
}