#1037 : 数字三角形（基础DP）

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题目链接

正向的计算，

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
//#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=211;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<=(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod  1000000007

int rewards[maxn][maxn];
int best[maxn][maxn];
int main()
{
    int n,m,i,j,t,k;
    int T;
    while(read(n))
    {
        memset(rewards,0,sizeof(rewards));
        for(i=1;i<=n;i++)
            for(j=1;j<=i;j++)
                scanf("%d",&rewards[i][j]);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=i;j++)
                best[i][j]=max(best[i-1][j-1],best[i-1][j])+rewards[i][j];
        }
        int ans=0;
        for(i=1;i<=n;i++)ans=max(ans,best[n][i]);
        printf("%d\n",ans);
    }

    return 0;
}
 

//逆向的计算

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
//#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=211;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<=(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod  1000000007

int rewards[maxn][maxn];
int best[maxn][maxn];
int main()
{
    int n,m,i,j,t,k;
    int T;
    while(read(n))
    {
        For(i,1,n)For(j,1,i)
        read(rewards[i][j]),rewards[i][i+1]=0;
        memset(best,0,sizeof(best));
        for(i=1;i<=n;i++)best[n][i]=rewards[n][i];
        for(i=n;i>=1;i--)
            for(j=1;j<=i;j++)
            best[i-1][j]+=max(best[i][j+1],best[i][j])+rewards[i-1][j];
        writeln(best[1][1]);
    }

    return 0;
}

#1037 : 数字三角形（基础DP）

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原文地址：http://blog.csdn.net/u013167299/article/details/45100735