UVA DP 入门专题

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674 - Coin Change17132

47.12%

5842

89.90%

10073734674Coin ChangeAcceptedC++2.0762012-05-04 11:09:02

状态转移方程：dp[j] = dp[j]+dp[j-cost[i]];

#include<iostream>#include<cstdio>#include<cstring>#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;}

10131 - Is Bigger Smarter?16114

37.77%

4752

84.85%

1007442610131Is Bigger Smarter?AcceptedC++0.0162012-05-04 14:45:20

先按重量大小排序，然后求IQ的最长递减子序列

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<stack>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<b.iq;return a.weight<b.weight;}class Ans{public:int num;int pre;}ans[1005];int main(){  //freopen("input.txt","r",stdin);int index=0,i,j;while(scanf("%d %d",&elephant[index].weight,&elephant[index].iq)!=EOF)elephant[index].number=index+1, ++index;sort(elephant,elephant+index,cmp);for(i=0; i<index; ++i){ans[i].num=1, ans[i].pre=i;}for(i=1; i<index; ++i){int maxNum=0,pre;for(j=0; j<i; ++j){if(elephant[j].weight<elephant[i].weight && elephant[j].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<index; ++i){if(ans[i].num>maxNum)maxNum=ans[i].num, maxPos=i;}printf("%d\n",maxNum);stack<int>st;while(maxNum--){st.push(elephant[maxPos].number);maxPos = ans[maxPos].pre;}while(!st.empty()){printf("%d\n",st.top());st.pop();}return 0;}

562 - Dividing coins16643

30.22%

3921

81.23%

10074473562Dividing coinsAcceptedC++0.0402012-05-04 15:03:51

01背包问题。

#include<cstdio>#include<cstring>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<m; ++i){scanf("%d",&value[i]);sum += value[i];}memset(dp,0,sizeof(dp));for(i=0; i<m; ++i){for(j=sum/2; j>=value[i]; --j)dp[j] = max(dp[j-value[i]]+value[i],dp[j]);}printf("%d\n",sum-dp[sum/2]*2);}return 0;}

10066 - The Twin Towers11414

44.45%

4465

86.58%

1007452510066The Twin TowersAcceptedC++0.0122012-05-04 15:26:08

最长公共子序列

#include<iostream>#include<cstdio>#include<cstring>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;// inputfor(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]);}}

10130 - SuperSale8092

49.68%

2865

92.11%

1007462010130SuperSaleAcceptedC++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<iostream>#include<cstdio>#include<cstring>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;}

10192 - Vacation12967

35.15%

3953

88.24%

1007471610192VacationAcceptedC++0.0082012-05-04 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<iostream>#include<cstdio>#include<cstring>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;}

357 - Let Me Count The Ways20522

27.54%

5907

66.84%

10074863357Let Me Count The WaysAcceptedC++0.7082012-05-04 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<iostream>#include<cstdio>#include<cstring>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;}

10465 - Homer Simpson8983

35.24%

2360

82.20%

1007502710465Homer SimpsonAcceptedC++0.2602012-05-04 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<iostream>#include<cstdio>#include<cstring>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<t)++cnt2;}printf("%d\n",cnt1+cnt2);}else{while(cnt1*m+cnt2*n!=dp[t]){--cnt1;while(cnt1*m+cnt2*n<dp[t])++cnt2;}printf("%d %d\n",cnt1+cnt2,t-dp[t]);}}return 0;}

// 版本2#include<iostream>#include<cstdio>#include<cstring>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]);elseprintf("%d %d\n",num[t],t-dp[t]);}return 0;}

—— 生命的意义，在于赋予它意义。

原创 http://blog.csdn.net/shuangde800 ， By D_Double