【codechef】Lazy Players

来源：互联网发布：新域名多长时间生效编辑：程序博客网时间：2024/05/21 20:21

Concerned with the fitness levels of the players in the National Team, the coach decides to carry out a running drill in the next training session. So, he sets up checkpoints in the training ground at different positions.

After evaluating each player’s fitness, the coach asks each player to cover a certain minimum distance while completing all checkpoints (Suppose a players starts at checkpoint 2, then he must cover all checkpoints and finish at checkpoint 2).

Players being lazy want to finish the drill in exact distance that the coach assigns them. Is it possible?

Input

Input description.

First line with two space separated integers N and L , the number of checkpoints and the minimum distance that coach asks the player to cover, respectively
Then N lines containing N space separated integers each. The jth integer on the ith line, d_ij, denotes the distance between checkpoint point i and j (d_ij for i != j, and d_ii = 0). For all 1 ≤ i, j, k ≤ N it is the case that d_ij = d_ji and d_ij ≤ d_ik + d_kj.

Output

For each test case, output a single line containing the answer i.e. POSSIBLE or IMPOSSIBLE.

Constraints

1 ≤ N ≤ 14
1 ≤ L ≤ 10¹⁴
1 ≤ d_ij ≤ L

Example

Input:5 150 2 3 3 22 0 3 2 33 3 0 2 23 2 2 0 32 3 2 3 0Output:POSSIBLE
http://www.codechef.com/problems/LAZY01
Dij表示i到j的距离。经过每个点后的总距离能否刚好等于15
#include<bits/stdc++.h>using namespace std;int main(){    int n,l,sum;    int a[14][14],i,j,k;    cin>>n>>l;    for(i=0;i<n;i++)        for(j=0;j<n;j++)            cin>>a[i][j];    for(i=0;i<n;i++){        sum=0;        for(j=0;j<n;j++){            for(k=0;k<n;k++){                if(sum<l){                    sum+=a[i][k]+a[k][j];  //有点类似floyd，不过很巧妙的方法                }            }        }        if(sum==l){            printf("POSSIBLE\n");            break;        }    }    if(i==n)        printf("IMPOSSIBLE\n");     return 0;}
#include <stdio.h>int main(void) {int a,b,i,j,max=0,sum=0;scanf("%d%d",&a,&b);int arr[a][a];for(i=0;i<a;i++){for(j=0;j<a;j++){scanf("%d",&arr[i][j]);}}for(i=0;i<a;i++){for(j=i+1;j<a;j++){if(arr[i][j]>max){max=arr[i][j];}}sum+=max;}if(sum>=b)printf("POSSIBLE");elseprintf("IMPOSSIBLE");return 0;}

0 0