Uva - 10131 - Is Bigger Smarter?
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题意:有n头大象,附来每只大象的重量和IQ,求按重量大且IQ小排成的最长子序列。
题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=114&problem=1072
——>>我觉得这种类型的题目,挺多都先排好其中一维,再给剩下的一维进行dp求解。
设d[i]表示从第i头大象出发的最长路径。
状态转移的方程:
d[i] = max(d[i], d[j]+1); i < j < m
结果有多种,所以不用纠结于样例中的输出,我取了字典序最小的输出。
#include <cstdio>#include <algorithm>using namespace std;const int maxn = 1000 + 10;const int INF = 214748364;struct ele{ int id; int vol; int iq; bool operator < (const ele& e) const { return vol < e.vol || (vol == e.vol && iq > e.iq); }}e[maxn];int d[maxn]; //dp数组int main(){ int vol, iq, i, j, m = 1; while(~scanf("%d%d", &vol, &iq)) { e[m].vol = vol; e[m].iq = iq; e[m].id = m++; } sort(e+1, e+m); //先排好其中一维,重量 d[m-1] = 1; for(i = m-2; i >= 1; i--) //dp求解 { d[i] = 1; for(j = m-1; j > i; j--) if(e[j].vol > e[i].vol && e[j].iq < e[i].iq) d[i] = max(d[i], d[j]+1); } int max_d = d[1], max_id = 1; for(i = 2; i < m; i++) //找最大值,且保证字典序最小 if(d[i] > max_d) { max_d = d[i]; max_id = i; } else if(d[i] == max_d) { if(e[i].id < e[max_id].id) max_id = i; } printf("%d\n", max_d); printf("%d\n", e[max_id].id); int temp = INF; for(i = max_id+1; i < m; i++) //输出路径,按字典序 { int id = INF; if(d[max_id] != d[i]+1) continue; while(i < m && d[max_id] == d[i]+1) { if(e[i].vol > e[max_id].vol && e[i].iq < e[max_id].iq) { id = min(id, e[i].id); temp = i; } i++; } max_id = temp; printf("%d\n", id); i--; } return 0;}
- UVa 10131 Is Bigger Smarter?
- UVa 10131 Is Bigger Smarter?
- UVA 10131 - Is Bigger Smarter
- uva 10131 Is Bigger Smarter?
- uva:10131 - Is Bigger Smarter?
- uva 10131 Is Bigger Smarter?
- UVA 10131 Is Bigger Smarter?
- UVA 10131 Is Bigger Smarter?
- UVa 10131 - Is Bigger Smarter?
- UVa 10131 - Is Bigger Smarter?
- Uva - 10131 - Is Bigger Smarter?
- UVa 10131 - Is Bigger Smarter?
- UVa 10131 - Is Bigger Smarter
- UVa 10131: Is Bigger Smarter?
- UVa:10131 Is Bigger Smarter?
- UVA - 10131 Is Bigger Smarter?
- UVa 10131 - Is Bigger Smarter?
- UVa 10131 - Is Bigger Smarter?
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