HEU The message

来源：互联网发布：用友erp软件价格编辑：程序博客网时间：2024/06/11 12:15

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] );    }}