HEU The message

囧啊，读错题了，一直在想为什么是DP。误以为当时间相同是可以一起删除。后来发现并非如此。用ans[i]记录前i条信息全部删除的最短时间，dele[i][i]表示删除第i条信息所需的时间，opt[a][b]是从a到b一起删除的时候所需的时间。用结构体表示到达时间（ arr ），删除所需时间（cost ）。

转移方程为 cnt = max ( ans[j], mess[i].arr ) + opt[j + 1][i] , ans[i] = min ( ans[i], cnt ) ( 0 <= j < i )

#include <cstdio>
#include <cstring>
#include <algorithm>
const int MAXX = 1 << 29;
using namespace std;struct Mess {int arr, cost;
}mess[505];int Case, ans[505], opt[505][505], m[505][505], l[505][505];void init ( ) {for ( int i = 1; i <= Case; ++i ) {l[i][i] = m[i][i] = opt[i][i] = mess[i].cost;ans[i] = MAXX;}for ( int i = 1; i < Case; ++i ) {for ( int j = i + 1; j <= Case; ++j ) {l[i][j] = min ( l[i][j - 1], mess[j].cost );m[i][j] = max ( m[i][j - 1], mess[j].cost );opt[i][j] = m[i][j] + l[i][j];}}
}void dp ( ) {init ( );for ( int i = 1; i <= Case; ++i )for ( int j = 0; j < i; ++j )if ( ans[i] >= max ( ans[j], mess[i].arr ) + opt[j + 1][i] ) ans[i] = max ( ans[j], mess[i].arr ) + opt[j + 1][i];
}bool cmp ( Mess a, Mess b ) {return a.arr < b.arr;
}int main ( ) {while ( scanf ( "%d", &Case ) != EOF ) {for ( int i = 1; i <= Case; ++i )scanf ( "%d%d", &mess[i].arr, &mess[i].cost );sort ( mess + 1, mess + Case + 1, cmp );dp ( );printf ( "%d\n", ans[Case] );}
}

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