POJ 1252 Euro Efficiency

Backpack or BFS

Italy is said to give you a few basic currencies, composed of 1~100 all currencies, using essentially the amount of money with the smallest.

Export usage probability. And the maximum amount of use.

It is possible to BFS.

Just remember to keep the array open. There may be 100 = 99+99-98 conditions.

A backpack is a complete backpack, the least likely to be added by how much.

Then make a 01 backpack to see if it can be reduced.

Backpack:

#include <cstdio> #include <cstring> #include <string> #include <queue> #include <algorithm > #include <map> #include <stack> #include <iostream> #include <list> #include <set># Include<cmath> #define INF 0x7fffffff#define EPS 1e-6#define LL long longusing namespace Std;int dp[10002];int value    [7];int Main () {int t;    scanf ("%d", &t);        while (t--) {int m=10001;        for (int i=0; i<6; i++) scanf ("%d", &value[i]);        for (int i=1; i<m; i++) dp[i]=10001;        dp[0]=0; for (int i=0, i<6; i++) {for (int j=value[i]; j+value[i]<m; J + +) {dp[j]=            Min (dp[j-value[i]]+1,dp[j]);//printf ("%d:%d==\n", j,dp[j]);                }} for (int i=0, i<6; i++) {for (int j=m-value[i]; j>0; j--) {      Dp[j]=min (Dp[j+value[i]]+1,dp[j]);//printf ("%d:%d==\n", j,dp[j]);      }} double ans=0;        int maxn=0;            for (int i=1; i<=100; i++) {//printf ("%d:%d\n", I,dp[i]);            Ans+=dp[i];        Maxn=max (Maxn,dp[i]);    } printf ("%.2f%d\n", ANS/100.0,MAXN); }}

BFS:

#include <cstdio> #include <cstring> #include <string> #include <queue> #include <algorithm > #include <map> #include <stack> #include <iostream> #include <list> #include <set># Include<cmath> #define INF 0x7fffffff#define EPS 1e-6#define LL long longusing namespace std;struct lx{int ans,lv ;};    int ans[2051];int value[7];void BFS () {queue<lx>q;    BOOL vis[2051];    memset (vis,0,sizeof (VIS));    LX now,tmp;    tmp.ans=0,tmp.lv=0;    Q.push (TMP);    Vis[0]=1;        while (!q.empty ()) {Tmp=q.front (); Q.pop ();        ans[tmp.ans]=tmp.lv;            for (int i=0;i<6;i++) {int num1=tmp.ans+value[i];            int num2=tmp.ans-value[i];            now.lv=tmp.lv+1;                if (!vis[num1]&&num1>0&&num1<2000) {vis[num1]=1;                NOW.ANS=NUM1;            Q.push (now);          } if (!vis[num2]&&num2>0&&num2<2000)  {vis[num2]=1;                now.ans=num2;            Q.push (now);    }}}}int Main () {int t;    scanf ("%d", &t);        while (t--) {for (int i=0; i<6; i++) scanf ("%d", &value[i]);        BFS ();        Double an=0;        int maxn=0;            for (int i=1;i<=100;i++) {an+=ans[i];        Maxn=max (Maxn,ans[i]);//printf ("%d:%d\n", i,ans[i]);    } printf ("%.2f%d\n", AN/100,MAXN); }}

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POJ 1252 Euro Efficiency