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  1. PO
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    <p>I'm not sure I get your solution idea, but I describe my AC solution:</p> <p>I'm using function with memorization, but you can re-write it using non-recurcive DP. </p> <p>Let's say we have our intervals in array </p> <p>pair a[100]; where a[i].first is interval begin and a[i].second is interval end.</p> <p>Sort this array by <em>begin</em> first (default behavior of stl sort algorithm with default pair comparator).</p> <p>Now imagine that we are 'putting' intervals one by one from beginning to end. </p> <p>let f(int x, int prev) return the number of ways to finish the filling if currently last interval is x and previous is 'prev'. </p> <p>we'll calculate it as follows:</p> <pre><code>int f(int x, int prev) { // if already calculated dp[x][prev], return it. Otherwise, calculate it if (dp[x][prev] != -1) { return dp[x][prev]; } if (a[x].second == m) { return dp[x][prev] = 1; // it means - X is last interval in day } else { dp[x][prev] = 0; for (int i = x + 1; i &lt; n; ++i) { // try to select next interval if (a[i].first &lt;= a[x].second &amp;&amp; // there must be not empty space after x interval a[i].second &gt; a[x].second &amp;&amp; // if this is false, the set won't be minimal - i interval is useless a[i].first &gt; a[x].first &amp;&amp; // if this is false, the set won't be minimal, x interval is useless a[prev].second &lt; a[i].first) { // if this is false, the set won't be minimal, x interval is useless. dp[x][prev] = (dp[x][prev] + f(i, x)) % 100000000; } } } return dp[x][prev]; } </code></pre> <p>After that we need to call this function for every pair of intervals, first of which start at 0 and second is connected with first:</p> <pre><code>for (int i = 0; i &lt; n; ++i) { if (a[i].first == 0) { for (int j = i + 1; j &lt; n; ++j) { if (a[j].first &gt; 0 &amp;&amp; // we don't need to start at 0 - in this case either i or j will be useless a[j].first &lt;= a[i].second &amp;&amp; // there must be no space after i interval a[j].second &gt; a[i].second) { // in opposite case j will be useless res = (res + f(j, i)) % 100000000; } } // also we need to check the case when we use only one interval: if (a[i].second == m) { res = (res + 1) % 100000000; } } } </code></pre> <p>After that we only need to print the res.</p>
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    1. POHelp in a DP problem
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    1. USOleksandr Kuvshynov
    1. USOleksandr Kuvshynov
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    1. COHow can it ever count anything? You only consider i values where a[i].first == 0, then you only consider j values where a[j].first > 0 and a[j].first <= a[i].first. Given the first comparison, the second is a contradiction and you will never call f(j, i)
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      1. USPeer Sommerlund
    2. COwhoops, made a mistake while formatting the code. Thanks, edited.
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      1. USOleksandr Kuvshynov
 

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