http://codeforces.com/gym/101617
题意：有n个金矿，每一天他都会生产金子，但是每天生产的金子会减少d，初始值生产是g，你只能拿当天生产的金子，金矿与金矿之间有路，需要花t天才能走过去。现在问你
最多最后能拿多少金子，另外不能在同一个金矿停留。
做法：珂学分析，DP，咋D呢，再次珂学分析，有一维肯定是第几天了，dp的值肯定是最后的金子数，那么还差一维啥呢，差一维第几个金矿，
dp[i][j]代表第i天到了第j个金矿的金子数。啊，对如闪电。
开始推方程dp[i][j]=max(dp[i][j],dp[i-t][v]+g[j]-d[j]*(i-1))
因为这金矿的状态一定是上一个金矿走过去的，花了t天，v是上一个金矿的标号，最后那一部分是金矿的生产值，仔细一想，不对，
dp[i][j]=max(dp[i][j],dp[i-t][v]+max(0,g[j]-d[j]*(i-1)))。
嗯！这样对了，就不解释了哈。
有了DP方程，建图，然后就更新ans啦。
于是:
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx)
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;
int n,m;
struct node
{
LL from,t;
};
LL dp[maxn][maxn];
LL g[maxn],d[maxn];
vector<node>G[maxn];
void solve()
{
s_2(n,m);
me(dp,-1);
FOR(1,n,i)
S_2(g[i],d[i]);
FOR(1,m,i)
{
LL a,b,t;
S_3(a,b,t);
G[a].pb({b,t});
G[b].pb({a,t});
}
dp[1][1]=g[1];
LL ans=0;
FOR(1,1000,i)
{
FOR(1,n,j)
for(auto u:G[j])
{
LL v=u.from;
LL time=u.t;
if(i-time>=1&&dp[i-time][v]!=-1)
dp[i][j]=max(dp[i][j],dp[i-time][v]+max(0ll,g[j]-d[j]*(i-1)));
if(dp[i][j]!=-1)
ans=max(ans,dp[i][j]);
}
}
print(ans);
}
int main()
{
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}A啦