UVA 10154 Weights and Measures
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大意不再赘述。
思路:一开始以为是DAG单源最长路径,后来WA,思考了很久,没有个头绪,主要是重量叠加问题比较难处理。
这里有几个问题比较难处理。
1、这里的乌龟是无序状态。
2、对于每一个乌龟来说重量是叠加的。
对于1,动态规划的本质就是有序的多阶段决策问题,根据题意,我们可以按照力量从小到大排一次序,这样方便规划方程的建立于求解。
对于2,由于乌龟的重量是叠加的,而且它们的力量是递增的,所以我们可以写出规划方程d[i][j] = min(d[i-1][j], d[i-1][j-1]+wi) 1 <= j <= i;
规划方程d[i][j]表示前i个乌龟能够叠加到j层的最小重量,这样即可保证重量是叠加的,而且子问题是最优解。
当前决策,选乌龟i叠加到j层或者不选乌龟i叠加到j层,不选乌龟i,那么d[i][j] = d[i-1][j],选了说明第j层的乌龟就是i,即d[i-1][j-1]+wi。
状态转移是取最优解,所以规划方程是:d[i][j] = min(d[i-1][j], d[i-1][j-1]+wi) 1 <= j <= i;
边界情况:d[i][0] = 0 (i = 1, 2, 3...n),表示叠加0层时,最小重量是0。
#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAXN = 5670;const int INF = 0x3f3f3f3f;struct node{int w, v;bool operator < (const node a) const{return v < a.v;}}A[MAXN];int n;int d[MAXN][MAXN];void init(){memset(d, INF, sizeof(d));for(int i = 0; i <= n; i++) d[i][0] = 0;}int dp(){for(int i = 1; i <= n; i++){for(int j = 1; j <= i; j++){d[i][j] = d[i-1][j];if(d[i-1][j-1] != INF && A[i].v >= d[i-1][j-1] + A[i].w){d[i][j] = min(d[i][j], d[i-1][j-1]+A[i].w);}}}for(int i = n; i >= 1; i--) if(d[n][i] != INF) return i;}void read_case(){n = 1;while(~scanf("%d%d", &A[n].w, &A[n].v) && A[n].w) n++;--n;sort(A+1, A+1+n);}void solve(){read_case();init();int ans = dp();printf("%d\n", ans);}int main(){solve();system("pause");return 0;}
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