BZOJ2612 [Poi2003]Sums

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任取一个物品，假设其体积为V，那么我们可以在模V的意义下做背包，f[i]表示对于模V得i的物品，当体积>=f[i]时能被表示出来

那么就可以跑最短路了

不妨取体积最小那个，dijkstra的话理论复杂度是(5000*50000)logn，但是跑的飞起

事实上我们可以取最大那个，然后令f[i]表示表示对于模V得i的物品，当体积>=f[i]*V时能被表示出来

这样的话边权都为0或者1，那么我们可以01bfs，理论复杂度(5000*50000)，但是跑不过

dijkstra:

#include<iostream>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>#include<iomanip>#include<cstdlib>#include<cstdio>#include<map>#include<bitset>#include<set>#include<stack>#include<vector>#include<queue>using namespace std;#define MAXN 50010#define MAXM 1010#define ll long long#define eps 1e-8#define MOD 1000000007#define INF 1000000000struct data{    int x;    int v;    data(){             }    data(int _x,int _v){        x=_x;        v=_v;    }    friend bool operator <(data x,data y){        return x.v>y.v;    }};int n;int a[MAXN];int mn;int f[MAXN];priority_queue<data>q;bool vis[MAXN];void dijkstra(){    int i,x;    memset(f,0x3f,sizeof(f));    f[0]=0;    q.push(data(0,0));    while(!q.empty()){        x=q.top().x;        q.pop();        if(vis[x]){            continue ;        }        vis[x]=1;        for(i=1;i<=n;i++){            if(f[x]+a[i]<f[(x+a[i])%mn]){                f[(x+a[i])%mn]=f[x]+a[i];                q.push(data((x+a[i])%mn,f[(x+a[i])%mn]));            }        }    }}int main(){    int i,j,x;    scanf("%d",&n);    for(i=1;i<=n;i++){        scanf("%d",&a[i]);    }    mn=a[1];    dijkstra();    scanf("%d",&n);    while(n--){        scanf("%d",&x);        printf(x>=f[x%mn]?"TAK\n":"NIE\n");    }    return 0;} /*325760141232 */

01bfs:

#include<iostream>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>#include<iomanip>#include<cstdlib>#include<cstdio>#include<map>#include<bitset>#include<set>#include<stack>#include<vector>#include<queue>using namespace std;#define MAXN 50010#define MAXM 1010#define ll long long#define eps 1e-8#define MOD 1000000007#define INF 1000000000int n;int a[MAXN];int mx;int f[MAXN];int q[MAXN],hd,tl;int Q[MAXN],HD,TL;void bfs(){int i,x;memset(f,0x3f,sizeof(f));f[0]=0;q[tl++]=0;while(hd!=tl){Q[TL++]=q[hd++];int thd=HD;while(HD!=TL){x=Q[HD++];for(i=1;i<=n;i++){if(f[(x+a[i])%mx]>INF){if(x+a[i]<mx){f[x+a[i]]=f[x];Q[TL++]=x+a[i];}}}}HD=thd;while(HD!=TL){x=Q[HD++];for(i=1;i<=n;i++){if(f[(x+a[i])%mx]>INF){if(x+a[i]>=mx){f[x+a[i]-mx]=f[x]+1;q[tl++]=x+a[i]-mx;}}}}}}int main(){int i,j,x;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);}mx=a[n];bfs();scanf("%d",&n);while(n--){scanf("%d",&x);printf(x/mx>=f[x%mx]?"TAK\n":"NIE\n");}return 0;}/*325760141232*/

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