Problem:
Score : 400 points

Problem Statement

There is a directed graph with N vertices and M edges.
The i-th edge (1≤i≤M) points from vertex a_i to vertex b_i, and has a weight c_i.
We will play the following single-player game using this graph and a piece.

Initially, the piece is placed at vertex 1, and the score of the player is set to 0.
The player can move the piece as follows:

When the piece is placed at vertex a_i, move the piece along the i-th edge to vertex b_i. After this move, the score of the player is increased by c_i.

The player can end the game only when the piece is placed at vertex N.
The given graph guarantees that it is possible to traverse from vertex 1 to vertex N.

When the player acts optimally to maximize the score at the end of the game, what will the score be?
If it is possible to increase the score indefinitely, print inf.

Constraints

2≤N≤1000

1≤M≤min(N(N-1),2000)

1≤a_i,b_i≤N (1≤i≤M)

a_i≠b_i (1≤i≤M)

a_i≠a_j or b_i≠b_j (1≤i<j≤M)

-10^9≤c_i≤10^9 (1≤i≤M)

c_i is an integer.

In the given graph, there exists a path from vertex 1 to vertex N.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the maximum possible score at the end of the game, if it is finite. If it is possible to increase the score indefinitely, print inf.

Sample Input 1

3 3
1 2 4
2 3 3
1 3 5

Sample Output 1

7

There are two ways to move the piece to vertex N=3:

vertex 1 → vertex 2 → vertex 3 : score 4+3=7

vertex 1 → vertex 3 : score 5

Thus, the maximum possible score at the end of the game is 7.

Sample Input 2

2 2
1 2 1
2 1 1

Sample Output 2

inf

It is possible to increase the score indefinitely by alternating between vertex 1 and 2.

Sample Input 3

6 5
1 2 -1000000000
2 3 -1000000000
3 4 -1000000000
4 5 -1000000000
5 6 -1000000000

Sample Output 3

-5000000000

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdlib.h>
#include <stdbool.h>
#define MOD 1000000007
#define END printf("\n");return 0;
#define QS09(how_data,data) qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_09);
#define QS90(how_data,data) qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_90);
#define I(a) long a;scanf("%ld",&a);
void input_array(long how_data,long *data);
void output_array(long how_data,long *data);
void input_array2(long first , long second , long data[][2]);
void format_array(long how_data ,long *data,long what);
long get_random(long min, long max);
long factorial(long n);
long fibonacci(long n);
int qsort_09(const int *sys1 , const int *sys2);
int qsort_90(const int *sys1 , const int *sys2);
long sel_max(long a , long b);
long sel_min(long a , long b);
long array_max(long how_data,long *data);
long array_min(long how_data,long *data);
long can_DP(long how_data,long *data,long how_can,bool *can);
long array_sum(long how_data,long *data);
long Leven_dist(char *now , char *target);
void prime_fact(long target, long *data);
long get_digit(long target);
long ncr(long n , long r);
long npr(long n , long r);
long nhr(long n , long r);
 
long loop1,loop2,loop3,loop4,loop5,i_temp;
char c_temp;

typedef struct{
    long from;
    long to;
    long omomi;
} KOUZOU;

int main(void){
    I(how_tyouten);I(how_hen);
    KOUZOU data[how_hen];
    for(loop1=0;loop1<how_hen;loop1++){
        scanf("%ld %ld %ld",&data[loop1].from,&data[loop1].to,&data[loop1].omomi);
    }
    
    //最短経路問題用のアルゴリズムを改造
    //ベルマンフォード法だと符号あってもいい？
    
    long max_score[1500];
    format_array(1500,max_score,-1000000000000000); //-infで初期化
    max_score[1]=0;
    
    //max_score[how_tyouten]をどれだけ大きくできるか？
    //無限ループをどうやって検知するか？
    bool flag=1;
    int keisankaisuu=0;
    while(flag){
        flag=0;
        for(loop1=0;loop1<how_hen;loop1++){
            if(max_score[data[loop1].from]+data[loop1].omomi > max_score[data[loop1].to]){
                max_score[data[loop1].to]=max_score[data[loop1].from]+data[loop1].omomi;
                flag=1;
            }
        }
        keisankaisuu++;
        if(keisankaisuu>how_tyouten+2000){
            printf("inf\n");
            END;
        }
    }
    
    printf("%ld\n",max_score[how_tyouten]);
    
    
    END;
}
 

 
void input_array(long how_data,long *data){
    long loop;
    for(loop=0;loop<how_data;loop++){
        scanf("%ld",&data[loop]);
    }
    return ;
}
void output_array(long how_data,long *data){
    long loop;
    for(loop=0;loop<how_data;loop++){
        printf("%ld ",data[loop]);
    }
    return ;
}
 
void input_array2(long first,long second,long data[][2]){
    long loopA,loopB;
    for(loopA=0;loopA<first;loopA++){
        for(loopB=0;loopB<second;loopB++){
            scanf("%ld",&data[loopA][loopB]);
        }
    }
    return ;
}
void format_array(long how_data ,long *data,long what){
    long loopA;
    for(loopA=0;loopA<how_data;loopA++){
        data[loopA]=what;
    }
    return ;
}
 
long get_random(long min, long max){   //指定の範囲から乱数を1つ返すやつ
    //srand((unsigned int)time(0));   //現在時刻で疑似乱数初期化
    rand();rand();rand();rand();    //乱数を空打ち
    return (rand()%(max+1-min))+min;
}
 
 
long factorial(long n){//n!のMOD10^9+7を返すやつ
    unsigned long long int ret=1;
    for(long i=1;i<=n;i++)ret=(ret*i)%1000000007;
    return (long)ret;
}
 
int qsort_09(const int *sys1 , const int *sys2){ //小さいのが上にくるやつ
    //qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_09);
    if(*sys1<*sys2){
        return -1;
    }else if(*sys1==*sys2){
        return 0;
    }else{
        return 1;
    }
}
 
int qsort_90(const int *sys1 , const int *sys2){ //大きいのが上にくるやつ
    //qsort(data,how_data,sizeof(long),(int (*)(const void *,const void *))qsort_90);
    if(*sys1<*sys2){
        return 1;
    }else if(*sys1==*sys2){
        return 0;
    }else{
        return -1;
    }
}
 
long fibonacci(long n){
    switch(n){
        case 0:return 0;
        case 1:return 1;
        default :return fibonacci(n-1)+fibonacci(n-2);
    }
}
 
long sel_max(long a,long b){
    if(a>b)return a;
    return b;
}
 
long sel_min(long a,long b){
    if(a>b)return b;
    return a;
}
 
long can_DP(long how_data,long *data,long how_can,bool *can){//Typical DP Contest A
    //data内で組み合わせられる和をcanに0 or 1で入れる
    //返り値はパターン数
    long loopA,loopB;
    long ret=0;
    for(loopA=0;loopA<how_can;loopA++){//初期化
        can[loopA]=0;
    }
    can[0]=1;//絶対にありえる
    for(loopA=0;loopA<how_data;loopA++){
        for(loopB=how_can-1;loopB>=0;loopB--){
            if(can[loopB]==1 && loopB+data[loopA]<how_can){
                can[loopB+data[loopA]]=1;
            }
        }
    }
    for(loopA=0;loopA<how_can;loopA++){
        if(can[loopA]==1){
            ret++;
        }
    }
    return ret;
}
long array_max(long how_data,long *data){
    long loop;
    long ret=data[0];
    for(loop=0;loop<how_data;loop++){
        if(ret<data[loop])ret=data[loop];
    }
    return ret;
}
long array_min(long how_data,long *data){
    long loop;
    long ret=data[0];
    for(loop=0;loop<how_data;loop++){
        if(ret>data[loop])ret=data[loop];
    }
    return ret;
}
long array_sum(long how_data,long *data){
    long ret=0;
    long loop;
    for(loop=0;loop<how_data;loop++){
        ret+=data[loop];
    }
    return ret;
}
long Leven_dist(char *now , char *target){
    long loopA,loopB;
    //レーベンシュタイン距離を求める
    // 参考文献
    // http://nw.tsuda.ac.jp/class/algoB/c13.html  (アルゴリズム理解)
    // http://d.hatena.ne.jp/ohnishiakira/20090809/1249845529  (実装)
    long len_now=strlen(now)+1;
    long len_target=strlen(target)+1;
    long d[len_now][len_target]; //計算用
    for(loopA=0;loopA<len_now;loopA++) d[loopA][0]=loopA;
    for(loopA=0;loopA<len_target;loopA++) d[0][loopA]=loopA;
    for(loopA=1;loopA<len_now;loopA++){
        for(loopB=1;loopB<len_target;loopB++){
            long cost=(now[loopA-1]==target[loopB-1] ? 0:1);
            d[loopA][loopB]=sel_min(sel_min(d[loopA-1][loopB]+1,d[loopA][loopB-1]+1),d[loopA-1][loopB-1]+cost);
        }
    }
    return d[len_now-1][len_target-1];
}
void prime_fact(long target, long *data){
    long loopB=0;
    long loopA=2;
    long moto_target=target;
    while(target!=1){
        loopA-=1;
        while(1){
            loopA++;
            if(loopA>=sqrt(moto_target)+100){
                data[loopB]=target;
                target=1;
                break;
            }
            if(target%loopA==0){
                data[loopB]=loopA;
                target/=loopA;
                loopB++;
                break;
            }
        }
    }
    return ;
}
long get_digit(long target){
    return (long)(log10(target)+1);
}
long ncr(long n , long r){
    //パスカルの三角形
    long loopA,loopB;
    long pascal[100][102]={{0}};
    pascal[1][0]=1;
    pascal[1][1]=1;
    for(loopA=2;loopA<100;loopA++){
        pascal[loopA][0]=1;
        for(loopB=1;loopB<loopA;loopB++){
            pascal[loopA][loopB]=pascal[loopA-1][loopB-1]+pascal[loopA-1][loopB];
        }
        pascal[loopA][loopA]=1;
    }
    return pascal[n][r];
}
long npr(long n, long r){
    return ncr(n,r)*factorial(r);
}
long nhr(long n , long r){
    return ncr(n+r-1,r);
}