UVA DP 入门专题-白红宇

UVA DP 入门专题

阅读量：4074 次

发布时间：2019-05-25

本文共 6313 字，大约阅读时间需要 21 分钟。

17132

47.12%

5842

89.90%

10073734

Accepted

C++

2.076

2012-05-04 11:09:02

状态转移方程：dp[j] = dp[j]+dp[j-cost[i]];

#include
    
     #include
     
      #include
      
       #define MAXN 7500using namespace std;int cost[6]={0,1,5,10,25,50};int dp[MAXN];int main(){  //freopen("input.txt","r",stdin);    int n,i,j;    while(scanf("%d",&n)!=EOF){        for(i=0; i<=n; ++i)            dp[i]=1;        for(i=2; i<=5; ++i){            for(j=cost[i]; j<=n; ++j)                dp[j]=dp[j]+dp[j-cost[i]];        }         printf("%d\n",dp[n]);    }    return 0;}

16114

37.77%

4752

84.85%

10074426

Accepted

C++

0.016

2012-05-04 14:45:20

先按重量大小排序，然后求IQ的最长递减子序列

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         using namespace std;class point{public:	int weight;	int iq;	int number;}elephant[1005];bool cmp(const point& a,const point& b){	if(a.weight==b.weight)		return a.iq
         
          elephant[i].iq && ans[j].num+1>maxNum) maxNum=ans[j].num+1, pre=j; } if(maxNum>ans[i].num){ ans[i].num = maxNum; ans[i].pre = pre; } } int maxPos=0, maxNum=1; for(i=1; i
          
           maxNum) maxNum=ans[i].num, maxPos=i; } printf("%d\n",maxNum); stack
           
            st; while(maxNum--){ st.push(elephant[maxPos].number); maxPos = ans[maxPos].pre; } while(!st.empty()){ printf("%d\n",st.top()); st.pop(); } return 0;}

16643

30.22%

3921

81.23%

10074473

Accepted

C++

0.040

2012-05-04 15:03:51

01背包问题。

#include
    
     #include
     
      int value[105],dp[25000];int max(int a,int b){ return a>b?a:b; }int main(){	//freopen("input.txt","r",stdin);	int cas, m, i, j;		scanf("%d",&cas);		while(cas--){		scanf("%d",&m);				int sum=0;		for(i=0; i
      
       =value[i]; --j)				dp[j] = max(dp[j-value[i]]+value[i],dp[j]);		}		printf("%d\n",sum-dp[sum/2]*2);	}	return 0;}

11414

44.45%

4465

86.58%

10074525

Accepted

C++

0.012

2012-05-04 15:26:08

最长公共子序列

#include
    
     #include
     
      #include
      
       using namespace std;int max(int a,int b){return a>b?a:b;}int dp[105][105];int tow1[105],tow2[105];int main(){//	freopen("input.txt","r",stdin);	int cas=1;	int N1,N2,i,j;	while(scanf("%d %d",&N1,&N2)!=EOF){		if(!N1&&!N2) break;				// input		for(i=1; i<=N1; ++i)			scanf("%d",&tow1[i]);		for(i=1; i<=N2; ++i)			scanf("%d",&tow2[i]);		printf("Twin Towers #%d\n",cas++);		memset(dp,0,sizeof(dp));			for(i=1; i<=N1; ++i){			for(j=1; j<=N2; ++j){				if(tow1[i]==tow2[j]) dp[i][j]=dp[i-1][j-1]+1;				else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);			}		}		printf("Number of Tiles : %d\n\n",dp[N1][N2]);	}}

8092

49.68%

2865

92.11%

10074620

Accepted

C++

0.216

分组01背包问题，分别计算每个家人能带的最大价值物品，再加起来求总和

#include
    
     #include
     
      #include
      
       using namespace std;int max(int a,int b){return a>b?a:b;}int value[1005],cost[1005],group[105];int dp[30000]; int main(){ //	freopen("input.txt","r",stdin);		int T,N,G,g,i,j;	scanf("%d",&T);	while(T--){		scanf("%d",&N);		for(i=1; i<=N; ++i)			scanf("%d %d",&value[i],&cost[i]);		scanf("%d",&G);		for(i=1; i<=G; ++i)			scanf("%d",&group[i]);		int maxValue=0;		for(g=1; g<=G; ++g){			memset(dp,0,sizeof(dp));			for(i=1; i<=N; ++i){				for(j=group[g]; j>=cost[i]; --j)					dp[j] = max(dp[j-cost[i]]+value[i],dp[j]);			}			maxValue += dp[group[g]];		}		printf("%d\n",maxValue);	}	return 0;}

12967

35.15%

3953

88.24%

10074716

Accepted

C++

0.008

2012-05-04 16:06:02

最长公共子序列，本来还以为要不要判断重复的，交了后发现直接A了

#include
    
     #include
     
      #include
      
       using namespace std;char str1[105],str2[105];int dp[105][105];int max(int a,int b){return a>b?a:b;} int main(int i, int j){ //	freopen("input.txt","r",stdin);		int cas=1;	while(gets(str1+1),gets(str2+1)){		if(str1[0]=='#') break;				int len1=strlen(str1+1);		int len2=strlen(str2+1);		memset(dp,0,sizeof(dp));			for(i=1; i<=len1; ++i){			for(j=1; j<=len2; ++j){				if(str1[i]==str2[j]) {					dp[i][j] = dp[i-1][j-1]+1;				}				else{					dp[i][j] = max(dp[i-1][j],dp[i][j-1]);				}			}		}		printf("Case #%d: you can visit at most %d cities.\n",cas++,dp[len1][len2]);	}	return 0;}

20522

27.54%

5907

66.84%

10074863

Accepted

C++

0.708

2012-05-04 16:55:01

和674一样。但是特别要注意结果等于1输出格式不一样，还有要用64位int。因为这个原因WA多次

#include
    
     #include
     
      #include
      
       using namespace std;int value[6]={0,1,5,10,25,50};long long dp[30005];long long max(long long a,long long b){return a>b?a:b;} int main(int i, int j){   //	freopen("input.txt","r",stdin);	int n;	while(scanf("%d",&n)==1){					for(i=0; i<=n; ++i)			dp[i]=1;		for(i=2; i<=5; ++i){			for(j=value[i]; j<=n; ++j)				dp[j] += dp[j-value[i]];		}		if(dp[n]==1)			printf("There is only %lld way to produce %d cents change.\n",dp[n],n);		else 			printf("There are %lld ways to produce %d cents change.\n",dp[n],n);	}	return 0;}

8983

35.24%

2360

82.20%

10075027

Accepted

C++

0.260

2012-05-04 17:49:56

这题的题意一开始还搞不懂，后来终于明白了，设Krusty-burgers有x个，Kwik-e-Mart有y个，

要使m*x+n*y<=t, 即m*x+n*y尽量接近t.如果不等于t，就要输出和t相差多少

#include
    
     #include
     
      #include
      
       using namespace std; int dp[10005],cost[2];int max(int a,int b){return a>b?a:b;}void swap(int &a,int &b){ int t=a; a=b; b=t; }int main(int i, int j){ //	freopen("input.txt","r",stdin);	int m,n,t;	while(scanf("%d %d %d",&m,&n,&t)!=EOF){		memset(dp,0,sizeof(dp));		if(m>n)			swap(m,n);		cost[0]=m, cost[1]=n;				for(i=0; i<2; ++i){			for(j=cost[i]; j<=t; ++j)				dp[j] = max(dp[j],dp[j-cost[i]]+cost[i]);		} 		int cnt1=dp[t]/m;		int cnt2=0;		if(dp[t]==t){			while(cnt1*m+cnt2*n!=t){				--cnt1;				while(cnt1*m+cnt2*n

// 版本2#include
    
     #include
     
      #include
      
       using namespace std; int dp[10005],num[10005],cost[2];int max(int a,int b){return a>b?a:b;}int main(int i, int j){//  	freopen("input.txt","r",stdin);	int m,n,t;	while(scanf("%d %d %d",&cost[0],&cost[1],&t)!=EOF){		memset(dp,0,sizeof(dp));		memset(num,0,sizeof(num));				for(i=0; i<2; ++i){			for(j=cost[i]; j<=t; ++j){				if(dp[j-cost[i]]+cost[i]>dp[j]){					dp[j] = dp[j-cost[i]]+cost[i];					num[j] = num[j-cost[i]]+1;				}				else if(dp[j-cost[i]]+cost[i]==dp[j] && num[j-cost[i]]+1>num[j]){					num[j] = num[j-cost[i]]+1;				}			}		}		if(dp[t]==t)			printf("%d\n",num[t]);		else			printf("%d %d\n",num[t],t-dp[t]);	}	return 0;}