hdu 4284 Travel floyd + 状压DP

http://acm.hdu.edu.cn/showproblem.php?pid=4284

题解：http://www.cnblogs.com/E-star/archive/2012/09/11/2680992.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll long long
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 107
using namespace std;
//freopen("din.txt","r",stdin);
struct node{
    int c,d;
    int id;
}p[N];
int n,m,moy,H;
int mp[N][N],dp[1<<15][N];

void init(){
    int i,j;
    for (i = 0; i <= n; ++i){
        for (j = 0; j <= n; ++j){
            mp[i][j] = (i != j)*inf;
        }
    }
}
void floyd(){
    int i,j,k;
    for (k = 1; k <= n; ++k){
        for (i = 1; i <= n; ++i){
            for (j = 1; j <= n; ++j){
                if (mp[i][k] != inf && mp[k][j] != inf && mp[i][j] > mp[i][k] + mp[k][j]){
                    mp[i][j] = mp[i][k] + mp[k][j];
                }
            }
        }
    }
}
int main(){
    //freopen("din.txt","r",stdin);
    int i,j,k,x,y,z;
    int t;
    scanf("%d",&t);
    while (t--){
        scanf("%d%d%d",&n,&m,&moy);
        init();
        for (i = 0; i < m; ++i){
            scanf("%d%d%d",&x,&y,&z);
            if (mp[x][y] > z)//注意这里有重边
            mp[x][y] = mp[y][x] = z;
        }
        floyd();//求出新距离
        scanf("%d",&H);
        for (i = 0; i < H; ++i)
        scanf("%d%d%d",&p[i].id,&p[i].c,&p[i].d);


        CL(dp,-1);
        //初始化1可到达的点
        for (i = 0; i < H; ++i){
            int v = p[i].id;
            if (moy - mp[1][v] - p[i].d >= 0){
                dp[1<<i][i] = moy - mp[1][v] - p[i].d + p[i].c;
            }
        }
        for (i = 1; i <= ((1<<H) - 1); ++i){
            for (j = 0; j < H; ++j){
                if (dp[i][j] == -1) continue;//对于以j结尾的i路线存在
                int u = p[j].id;
                for (k = 0; k < H; ++k){//查看j是否可到k
                    if ((i&(1<<k)) == 0){
                        int v = p[k].id;
                        if (dp[i][j] - mp[u][v] - p[k].d >= 0){//可到
                            int now = (i^(1<<k));
                            if (dp[now][k] == -1 || dp[now][k] < dp[i][j] - mp[u][v] - p[k].d + p[k].c)
                            dp[now][k] = dp[i][j] - mp[u][v] - p[k].d + p[k].c;
                        }
                    }
                }
            }
        }

        bool flag = false ;
        for (i = 0; i < H; ++i){
            if (dp[((1<<H) - 1)][i] - mp[p[i].id][1] >= 0){
                flag = true;
                break;
            }
        }
        if (flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}