HDU 1160 FatMouse's Speed DP題解
阿新 • • 發佈:2017-06-18
記錄 題解 tracking () mod opera find don incr
本題就先排序老鼠的重量,然後查找老鼠的速度的最長遞增子序列,只是由於須要按原來的標號輸出,故此須要使用struct把三個信息打包起來。
查找最長遞增子序列使用動態規劃法。主要的一維動態規劃法了。
記錄路徑:僅僅須要記錄後繼標號,就能夠逐個輸出了。
#include <stdio.h> #include <algorithm> using namespace std; const int MAX_N = 1005; struct MouseSpeed { int id, w, s; bool operator<(const MouseSpeed &ms) const { return w < ms.w; } }; int N; MouseSpeed msd[MAX_N]; int post[MAX_N], tbl[MAX_N]; int main() { N = 0; while (~scanf("%d %d", &msd[N].w, &msd[N].s)) { msd[N].id = N+1; ++N; } sort(msd, msd+N); //fullfill condition 1: weight increase fill(tbl, tbl+N, 1);//initialize dynamic table post[N-1] = N-1; for (int i = N-2; i >= 0; i--) { post[i] = i; //as the print out terminate term for (int j = i+1; j < N; j++) { if (msd[i].s>msd[j].s && msd[i].w!=msd[j].w && tbl[i]<tbl[j]+1) {//strictly increase, so don‘t forget msd[i].w must < msd[j].w tbl[i] = tbl[j]+1;//update longest subsequence post[i] = j;//record the post } } } int id = 0, maxSeq = 0; for (int i = 0; i < N; i++)//find the max sequence starting point { if (maxSeq < tbl[i]) { id = i; maxSeq = tbl[i]; } } printf("%d\n", maxSeq); printf("%d\n", msd[id].id); while (id != post[id]) { id = post[id]; printf("%d\n", msd[id].id);//print out the original indices } return 0; }
HDU 1160 FatMouse's Speed DP題解