poj_3176

源地址：http://poj.org/problem?id=3176

题目大意就是一个三角形，从顶部选一条路到低端，使得这条经过的路上的权值和最大，路只能是从上一层的j到下一层的j和j+1。

经典的dp，从底往上走一遍就行。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m;#define Mod 1000000007#define N 510#define M 1000100int dp[N][N];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    while(sf(n)!=EOF){    for(int i=1;i<=n;i++){    for(int j=1;j<=i;j++)    sf(dp[i][j]);    }    for(int i=n-1;i>=1;i--){    for(int j=1;j<=i;j++){    dp[i][j] = Max(dp[i+1][j],dp[i+1][j+1])+dp[i][j]; //选择左右两边大的那个加上去即可    }    }    printf("%d\n",dp[1][1]);    }return 0;}