hdu4781Assignment For Princess（2013 成都）构造题

参考：http://blog.csdn.net/diary_yang/article/details/16368039

首先所有回路的边值和都为%3==0,至少有一个回路

则先构造一个大的回路，将所有的点包含在内，n点，n边，由于n+3<=m,必定可以构造出来，此时此图是满足要求的

然后再此图的基础上加边，并保证仍然满足要求即可。若找不到给定值的边，则print-1

ps：

（1）题目10 <= N <= 80, N+3 <= M <= N²/7

可是却给了个n=6，m=8的样例，明显错误的

（2）sum+i%3==0出错了，（sum+i）% 3 == 0，加括号才行

//#pragma comment(linker, "/STACK:1024000000,1024000000")#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>using namespace std;#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FD(i, b, a) for(int i = (b); i >= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(a, v) memset(a, v, sizeof(a))#define PB push_back#define MP make_pairtypedef long long LL;const int INF = 0x3f3f3f3f;const int maxn = 100;int n, m;int dis[maxn][maxn], g[maxn][maxn];vector<pair<int, int> >E[maxn];bool vis[maxn * maxn];int dfs(int u, int pos, int fa, int val){    if (u == pos) return val;    REP(r, E[u].size())    {        int v = E[u][r].first, d = E[u][r].second;        if (v != fa) return dfs(v, pos, u, val + d);    }}void pre(){    FE(i, 1, n) FE(j, 1, n)        g[i][j] = dfs(i, j, -1, 0);}bool check(int x){    FE(i, 1, n) FE(j, 1, n)    {        if (i != j)        if (!dis[i][j] && !dis[j][i] && g[i][j] % 3 == x % 3)        {            dis[i][j] = x;            return true;        }    }    return false;}bool solve(){    FE(i, 1, m)    {        if (!vis[i] && !check(i))            return false;    }    return true;}int main (){    int T;    scanf("%d",&T);    int nc = 1;    while (T--)    {        scanf("%d%d", &n, &m);        CLR(vis, 0);        CLR(dis, 0);        REP(i, maxn) E[i].clear();        int sum = 0;        FE(i, 1, n - 1)        {            vis[i] = 1;            E[i].PB(MP(i + 1, i));            dis[i][i + 1] = i;            sum += i;        }        FE(i, n, n + 3)        {            if (!vis[i] && sum + i % 3 == 0)            {                vis[i] = 1;                E[n].PB(MP(1, i));                dis[n][1] = i;                break;            }        }        pre();        printf("Case #%d:\n", nc++);        if (solve())        {            FE(i, 1, n) FE(j, 1, n)            {                if (dis[i][j])                {                    printf("%d %d %d\n", i, j, dis[i][j]);                }            }        }        else printf("-1\n");    }    return 0;}