UVA - 11578（背包类dp）

来源：互联网发布：陆维网络图软件编辑：程序博客网时间：2024/06/05 18:09

首先，要对所有的要求，先按时间排序。然后只需对每一步若当前的要求和下一个时间相同，那么把他们两个同时安排，不同，则只安排当下的这个要求，选择有两种1,2

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <set>#include <algorithm>#include <vector>#include <queue>#include <map>using namespace std;typedef unsigned long long LLU;#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)#define rep1(i,n) for(int (i)=1;(i)<=(n);(i)++)const int maxn = 20100;int d[maxn][6][6],n,b[maxn];struct node{  int t,inc,id;  int i,j,k,s1,s2;  node(int i=0,int j=0,int k=0,int s1=0,int s2=0):      i(i),j(j),k(k),s1(s1),s2(s2){}  bool operator<(const node& a)const{    return t<a.t||(t==a.t&&inc<a.inc);  }}a[maxn],pa[maxn][6][6];void print_ans(int i,int j,int k){  if(i == n+1) return ;  node& te = pa[i][j][k];  if(te.i == i+2){     b[a[i].id] = te.s1; b[a[i+1].id] = te.s2;  }  else b[a[i].id] = te.s1;  print_ans(te.i,te.j,te.k);}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        memset(a,0,sizeof(a));        int res = 15*n;        rep1(i,n){          scanf("%d %d",&a[i].t,&a[i].inc);          a[i].inc/=10;          a[i].id = i;        }        sort(a+1,a+1+n);        for(int i=n+1;i>=1;i--)         rep1(j,5) rep1(k,5){            if(i == n+1){             d[i][j][k] = (j+k-2)*10;             continue;            }            d[i][j][k] = 1e9;            int& ans = d[i][j][k];            node& path=pa[i][j][k];            if(i+1<=n&&a[i+1].t==a[i].t){               int ans1 = d[i+2][a[i].inc][a[i+1].inc]+(abs(a[i].inc-j)+abs(a[i+1].inc-k))*10;               int ans2 = d[i+2][a[i+1].inc][a[i].inc]+(abs(a[i+1].inc-j)+abs(a[i].inc-k))*10;               if(ans1 <= ans2){                  ans = ans1; path=node(i+2,a[i].inc,a[i+1].inc,1,2);               }               else {                  ans = ans2; path=node(i+2,a[i+1].inc,a[i].inc,2,1);               }            }            else{                 int ans1 = d[i+1][a[i].inc][k]+abs(a[i].inc-j)*10;                 int ans2 = d[i+1][j][a[i].inc]+abs(a[i].inc-k)*10;               if(ans1<=ans2){                  path = node(i+1,a[i].inc,k,1); ans = ans1;              }              else {                 path =  node(i+1,j,a[i].inc,2); ans = ans2;              }            }        }        printf("%d\n",d[1][1][1]+res);        print_ans(1,1,1);        rep1(i,n) {         printf("%d\n",b[i]);       }    }    return 0;}/*132 402 501 40*/

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