HDU 4284 Travel(12年天津online floyd + tsp状态DP)

参考：http://blog.youkuaiyun.com/acm_cxlove/article/details/7963286

tsp的一些讲解：http://blog.youkuaiyun.com/gfaiswl/article/details/4749713

floyd + tsp

（1）如果没有点0，将其加入，并设c[],d[]，都为0即可

（2）dp[i][j]表示从0开始经过集合j中的点最后在i点的最大钱数

（3）最后判断，dp[i][ALL]表示遍历了所有的点后，位于i点的最大钱数，则判dp[i][ALL]和dis[num[i]][0]的大小关系即可

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
#include <bitset>
using namespace std;

//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
typedef long long LL;
const int INF = 1000000007;
const int MOD = 1000000007;
const double eps = 1e-10;
const int MAXN = 31000;

int ALL;
int dis[110][110];
int dp[20][1 << 17];
int c[20], d[20];
int num[20];
int n;
void floyd()
{
    REP(k, n) REP(i, n) REP(j, n)
    {
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
    }
}
int main()
{
    int m, h;
    int money;
    int x, y, z;
    int T;
    RI(T);
    while (T--)
    {
        RIII(n, m, money);
        REP(i, n + 1) REP(j, n + 1)
            dis[i][j] = i == j ? 0 : INF;
        while (m--)
        {
            RIII(x, y, z);
            x--; y--;
            dis[x][y] = min(dis[x][y], z);
            dis[y][x] = dis[x][y];
        }
        floyd();

        int pos = -1;
        RI(h);
        REP(i, h)
        {
            RIII(num[i], d[i], c[i]);
            num[i]--;
            if (num[i] == 0)
                pos = i;
        }
        if (pos == -1)
        {
            num[h] = 0; c[h] = 0; d[h] = 0;
            pos = h++;
        }

        ALL = (1 << h) - 1;
        CLR(dp, -1);///
        REP(i, h)
        if (money - c[i] - dis[num[i]][0] >= 0)
        dp[i][1 << i] = money - c[i] + d[i] - dis[num[i]][0];

//        if (money - c[pos] >= 0) dp[pos][1 << pos] = money - c[pos] + d[pos];
//        dp[pos][0] = money;///初始状态

        REP(ss, ALL + 1)
        REP(i, h)
        {
            if (dp[i][ss] == -1) continue;
            REP(j, h)
            {
//                if (i != j && (ss & (1 << j))) continue;
                if (ss & (1 << j)) continue;
                if (dp[i][ss] >= dis[num[i]][num[j]] + c[j])
                    dp[j][ss | (1 << j)] = max(dp[j][ss | (1 << j)], dp[i][ss] - dis[num[i]][num[j]] - c[j] + d[j]);
            }
        }

        int ans = 0;
        REP(i, h)
        {
            if (dp[i][ALL] >= dis[num[i]][0])
            {
                ans = 1;
                break;
            }
        }
        if (ans)
        puts("YES");
        else
        puts("NO");
    }
    return 0;
}