HDU 5305 Friends (second DFS + pruning for 2015 + schools)

FriendsTime limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 552 Accepted Submission (s): 253


Problem Descriptionthere is People and Pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline F Riends (mostly using face-to-face communication). However, everyone in these People wants to having the same number of online and offline friends (i.e. If one person has Onine friends, he or she must have Offline friends too, but different people can has different number of online or offline friends). Determine how many ways there is to satisfy their requirements.

Inputthe first line of the input was a single integer , indicating the number of testcases.

For each testcase, the first line contains the integers < Span class= "Mrow" id= "mathjax-span-26" style= "" > and , indicating the number of people and the number of pairs of friends, respectively. Each of the next Lines contains, numbers and , which mean and Is friends. It's Guaranteed that And every friend relationship would appear at the most once.

Outputfor each testcase, print one number indicating the answer.
Sample Input

23 31 22 33 14 41 22 33 44 1

Sample Output

02

Source2015 multi-university Training Contest 2 problem Solving ideas:
Notice that the data range is small, so brute force search + pruning.

#include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include < algorithm>using namespace Std;const int maxn = 30;int X[MAXN],Y[MAXN], S[MAXN], S1[MAXN], S2[MAXN], G[maxn][maxn];int n , M;int ans;void dfs (int dep) {if (dep > m) {for (int i=1;i<=n;i++) {if (S1[i]! = S2[i        ]) return;        } ans++;    return;    } int u = x[dep], v = y[dep];        if (S1[u] < S[u]/2 && s1[v] < S[v]/2) {s1[u]++; s1[v]++;        DFS (DEP + 1); s1[u]--;    s1[v]--;        } if (S2[u] < S[u]/2 && s2[v] < S[v]/2) {s2[u]++; s2[v]++;        DFS (DEP + 1); s2[u]--;    s2[v]--;    }}int Main () {int T;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n, &m);        memset (g, 0, sizeof (g));            for (int i=1; i<=m; i++) {scanf ("%d%d", &x[i], &y[i]); G[x[i]][y[i]] = G[y[i]][x[i]]= 1;        } int flag = 1;            for (int i=1; i<=n; i++) {s[i] = s1[i] = S2[i] = 0;            for (int j=1; j<=n; J + +) {if (i! = J && G[i][j] = = 1) s[i]++;        } if (S[i] & 1) flag = 0;            } if (!flag) {printf ("0\n");        Continue        } ans = 0;        DFS (1);    printf ("%d\n", ans); }}

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HDU 5305 Friends (second DFS + pruning for 2015 + schools)