codeforce 416C Booking System

原题地址

题意

一家餐馆有K张桌子，每张桌子有最大容量ri；

有N份订单，每份订单有人数和钱

要求每份订单的人必须坐在一张桌子上，否则无法接受该订单

问怎样使钱数最大

#include<bits/stdc++.h>using namespace std;const int maxn=1000;struct Request{int p;int c;int id;};struct P{int r;int id;};Request re[maxn+5];P p[maxn+5];int dp[maxn+5][maxn+5];bool cmp(const Request &a,const Request &b){return a.c>b.c;}bool cmp2(const P &a,const P &b){return a.r>b.r;}void print_ans(int n,int i,int j,int m,int s);int main(void){#ifdef ex freopen ("in.txt","r",stdin);#endifint n,k;int m,s;scanf("%d",&n);for (int i=1;i<=n;++i){scanf("%d%d",&re[i].c,&re[i].p);re[i].id=i;}scanf("%d",&k);for (int i=1;i<=k;++i){scanf("%d",&p[i].r);p[i].id=i;}sort(p+1,p+k+1,cmp2);sort(re+1,re+n+1,cmp);re[0]=(Request){100000,1000000,n+1};for (int i=n;i>=0;--i){for (int j=k;j>=1;--j){if (p[j].r>=re[i].c){dp[i][j]=max(dp[i+1][j],dp[i+1][j+1]+re[i].p);}else{dp[i][j]=dp[i+1][j];}}}s=dp[0][1];print_ans(n,0,1,0,s);}void print_ans(int n,int i,int j,int m,int s){if (i==n+1){printf("%d %d\n",m,s);return;}else if (dp[i][j]==dp[i+1][j]){print_ans(n,i+1,j,m,s);}else if (dp[i][j]==dp[i+1][j+1]+re[i].p){print_ans(n,i+1,j+1,m+1,s);printf("%d %d\n",re[i].id,p[j].id);}}

题解

DP

为了使DP的决策有序化，先把订单和桌子容量排序

然后转移就好了

问题出在路径打印上，一是打印的路径编号是排序后的而非原来的导致WA，二是打印函数两个else if 的次序不同会导致WA，还没想清楚为什么

下面的代码用回朔法打印路径，比较好写一点，智障的递归水平非常糟糕啊……

#include<bits/stdc++.h>using namespace std;const int maxn=1000;struct Request{int p;int c;int id;};struct P{int r;int id;};struct Cood{int i;int j;};Request re[maxn+5];P p[maxn+5];Cood t[maxn+5][maxn+5];Cood ans[maxn+5];int dp[maxn+5][maxn+5];bool cmp(const Request &a,const Request &b){return a.c>b.c;}bool cmp2(const P &a,const P &b){return a.r>b.r;}void print_ans(int n,int i,int j,int m,int s);int main(void){#ifdef ex freopen ("in.txt","r",stdin);#endifint n,k;int m,s;scanf("%d",&n);for (int i=1;i<=n;++i){scanf("%d%d",&re[i].c,&re[i].p);re[i].id=i;}scanf("%d",&k);for (int i=1;i<=k;++i){scanf("%d",&p[i].r);p[i].id=i;}sort(p+1,p+k+1,cmp2);sort(re+1,re+n+1,cmp);re[0]=(Request){100000,1000000,n+1};for (int i=n;i>=0;--i){for (int j=k;j>=1;--j){if (p[j].r>=re[i].c){if (dp[i+1][j+1]+re[i].p>dp[i+1][j]){dp[i][j]=dp[i+1][j+1]+re[i].p;t[i][j]=(Cood){i+1,j+1};}else{dp[i][j]=dp[i+1][j];t[i][j]=(Cood){i+1,j};}}else{dp[i][j]=dp[i+1][j];t[i][j]=(Cood){i+1,j};}}}s=dp[0][1];m=0;Cood st=t[0][1];while (1){int i=st.i;int j=st.j;st=t[i][j];if (st.j-j==1){++m;ans[m]=(Cood){re[i].id,p[j].id};}if (i==n) break;}printf("%d %d\n",m,s);for (int i=1;i<=m;++i){printf("%d %d\n",ans[i].i,ans[i].j);}}