CF 400B - Inna and New Matrix of Candies
2014-03-16 19:36
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英语是硬伤。题目大意:每次能让所有不在糖果上的小人一直往右移动直到:
1.到达最右边 2.到达糖果
求所有棋子到达糖果的最小步数,没有则-1。
思路:暴力模拟
不断找到最短的d(minD),往右移动minD步(即所有d减小minD),直到所有d都为0
1.到达最右边 2.到达糖果
求所有棋子到达糖果的最小步数,没有则-1。
思路:暴力模拟
不断找到最短的d(minD),往右移动minD步(即所有d减小minD),直到所有d都为0
#include<stdio.h> const int N=1002; char mp ; int g ,s ,d ; int min(int a,int b){ return a>b?b:a; } int main(){ int n,m; scanf("%d %d",&n,&m); for(int i=0;i<n;i++) scanf("%s",mp[i]); int lose=0; for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ if(mp[i][j]=='G') g[i]=j; else if(mp[i][j]=='S') s[i]=j; } if(g[i]>s[i]) lose=1; d[i]=s[i]-g[i]; } if(lose){ printf("-1\n"); return 0; } int ans=0; for(;;){ int minD=N; for(int i=0;i<n;i++) if(d[i]>0) minD=min(minD,d[i]); if(minD==N) break; //printf("%d\n",minD); for(int i=0;i<n;i++) if(d[i]>0) d[i]-=minD; ans++; } printf("%d\n",ans); return 0; }
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