結果

問題 No.92 逃走経路
ユーザー TLwiegehtt
提出日時 2015-07-17 01:22:48
言語 C90
(gcc 12.3.0)
結果
AC  
実行時間 4 ms / 5,000 ms
コード長 1,322 bytes
コンパイル時間 117 ms
コンパイル使用メモリ 24,064 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-07-08 08:18:57
合計ジャッジ時間 755 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 18
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.c: In function ‘main’:
main.c:26:9: warning: ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   26 |         scanf("%d %d %d", &N, &M, &K);
      |         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
main.c:30:17: warning: ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   30 |                 scanf("%d %d %d", &a, &b, &c);
      |                 ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~
main.c:45:17: warning: ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   45 |                 scanf("%d", &toll[i]);
      |                 ^~~~~~~~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <math.h>
#include <string.h>
#include <stdio.h>
#define NMAX (100)
#define TMAX (1000)
#define INF (99999999)
int toll[TMAX+5];
typedef struct{
int lnum;
int loadA[TMAX+5];
int loadB[TMAX+5];
}TOWN;
TOWN town[NMAX+3];
int main(void){
int nowTown[NMAX+5];
int i,j,k, N, M, K;
int cnt=0;
int minTown = INF;
scanf("%d %d %d", &N, &M, &K);
for(i=0;i<M;i++){
int a,b,c;
int ltmp;
scanf("%d %d %d", &a, &b, &c);
ltmp = town[a].lnum;
town[a].loadA[ltmp] = b;
town[a].loadB[ltmp] = c;
town[a].lnum += 1;
ltmp = town[b].lnum;
town[b].loadA[ltmp] = a;
town[b].loadB[ltmp] = c;
town[b].lnum += 1;
}
for(i=0;i<K;i++){
scanf("%d", &toll[i]);
toll[i+1] = 0;
}
for(i=0;i<=N;i++){ nowTown[i] = 1;}
for(i=0;i<K;i++){
int ntTmp[NMAX+5]={0};
for(j=1;j<=N;j++){
int ltmp = town[j].lnum;
if(nowTown[j] != 1){continue;}
for(k=0;k<ltmp;k++){
if( town[j].loadB[k] == toll[i] ){
ntTmp[ town[j].loadA[k] ] = 1;
}
}
}
for(j=0;j<=N;j++){nowTown[j] = ntTmp[j];}
}
for(i=1;i<=N;i++){
if(nowTown[i] == 1){
if(i < minTown){
minTown = i;
}
cnt++;
}
}
printf("%d\n", cnt);
printf("%d", minTown);
for(i=minTown+1;i<=N;i++){
if(nowTown[i] == 1){
printf(" %d", i);
}
}
printf("\n");
return 0;
}
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