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题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1108
解题报告:
1、首先按照weight从小到大排列,weight相同的按照speed从大到小排列;
2、Count[i]表示到第i个老鼠时,所求的最长“速度递减”子序列的长度;
3、path[i]=j是题目的关键,记录在Count[i]=Count[j]时,即最长“速度递减”子序列最后一个老鼠的前一只老鼠的位置
4、递归输出id
void output(int path[],pos) { if(pos==0) return ; output(path,path[pos]); printf("%d\n",mice[pos].id); }
Sources Code:
#include <cstdio> #include <stdlib.h> #include <algorithm> #define INF 0x3f3f3f3f using namespace std; int n;///n只老鼠 int Count[1001]= {0}; ///count[i]表示构造到第i个老鼠时,序列的最大长度 int path[1001]= {0}; ///path[i]表示构造到第i个老鼠时的前一个老鼠(前驱) struct mouse { int weight; int speed; int id; } mice[1001]; int cmp(const void *a,const void *b) { struct mouse *p1=(mouse *)a; struct mouse *p2=(mouse *)b; if(p1->weight==p2->weight) { if(p1->speed>p2->speed) return p2->speed-p1->speed; else return p1->speed-p2->speed; } else return p1->weight-p2->weight; } void output(int path[],int pos) { if(pos==0) return ; output(path,path[pos]); printf("%d\n",mice[pos].id); } int main() { n=0; int i=0,j=0; while(scanf("%d%d",&mice[++i].weight,&mice[++j].speed)!=EOF) { n++; mice[n].id=n; } qsort(mice+1,n,sizeof(mice[0]),cmp); Count[1]=1; for(int i=2; i<=n; i++) { for(int j=1; j<i; j++) { if(mice[i].weight>mice[j].weight&&mice[i].speed<mice[j].speed) { if(Count[i]<Count[j]) { Count[i]=Count[j]; path[i]=j; } } } Count[i]++; } int _max=0; int pos; for(int i=1; i<=n; i++) { if(Count[i]>_max) { _max=Count[i]; pos=i; } } printf("%d\n",_max); output(path,pos); }
动态规划(DP),类似LIS,FatMouse's Speed
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原文地址:http://www.cnblogs.com/TreeDream/p/5263661.html