uva_10154 - Weights and Measures (普通DP)

这题阶段性比较明显，可以事实上气力越大的乌龟放在越下面越有优势， So，阶段就是按照气力排序状态： dp[i][j] 表示 前i个乌龟可以堆积成高度为j个最小重量状态转移: dp[i][j] = min(dp[i][j-1], dp[i-1][j-1]+w[i](s[i] >= w[i]+dp[i-1][j-1]))；#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define MAXN    5700#define INF     0x3f3f3f3ftypedef struct NODE_ {        int w, s;}NODE;NODE turtle[MAXN];int dp[MAXN][MAXN];int cmp(const NODE &a, const NODE &b){        if( a.s == b.s ) {                return a.w < b.w;        }        return a.s < b.s;}int dynamic_pro(const int &n){        sort(turtle+1, turtle+n+1, cmp);        for(int i = 0; i <= n; i ++) {                dp[0][i] = dp[i][0] = 0;        }        for(int i = 1; i <= n; i ++) {                for(int j = 1; j <= i; j ++) {                        dp[i][j] = (!dp[i-1][j])? INF : dp[i-1][j];                        if( turtle[i].s >= dp[i-1][j-1]+turtle[i].w ) {                                dp[i][j] = min(dp[i][j], dp[i-1][j-1]+turtle[i].w);                        }                }        }        for(int i = n; i >= 1; i --) {                if( INF > dp[n][i] ) {                        return i;                }        }}int main(int argc, char const *argv[]){#ifndef ONLINE_JUDGE        freopen("test.in", "r", stdin);#endif        int idx(1);        while( ~scanf("%d %d", &turtle[idx].w, &turtle[idx].s) ) { idx ++; }        printf("%d\n", dynamic_pro(idx-1));        return 0;}