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题目好水,我是以座标DP的,最后一个点不一定要走,只要前面的点走过来可以超过后面的点也算的。dp[i]代表飞到座标i的最少步数 for(i=1;i+1<n;i++) { if(p[i].y<p[0].y)continue; lim=p[i].y*p[i].y-(p[i].y-p[0].y)*(p[i].y-p[0].y); for(j=0;j*j<=lim&&p[i].x-j>=p[0].x;j++) { if(dp[p[i].x-j]!=INF) { int st=p[i].x+j; if(st>p[n-1].x)st=p[n-1].x; if(dp[p[i].x-j]+1<dp[st]) { dp[st]=dp[p[i].x-j]+1; } } } } Followed by: Post your reply here: |
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