dfs

1、团队程序设计天梯赛-练习集-L3-001 凑零钱

参考：http://blog.csdn.net/chan_yeol/article/details/51362182

解题思路：

dfs + 剪枝

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <cmath>
#include <cctype>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const ull mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 1e4 + 10;
int N, M;
int num[maxn];
int sum = 0;
int ans[maxn];
int Count = 0;

bool dfs(int sum, int pos, int leave);

int main()
{
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
#endif // __AiR_H
scanf("%d%d", &N, &M);
for (int i = 0; i < N; ++i) {
scanf("%d", &num[i]);
sum += num[i];
}
sort(num, num+N);
bool flag = false;
for (int i = 0; i < N; ++i) {
if (dfs(num[i], i, sum)) {
flag = true;
printf("%d", num[i]);
for (int j = Count-1; j >= 0; --j) {
printf(" %d", ans[j]);
}
printf("\n");
break;
}
}
if (!flag) {
printf("No Solution\n");
}
return 0;
}

bool dfs(int sum, int pos, int leave)
{
if (sum > M || num[pos] > M) {
return false;
}
if (sum == M) {
return true;
}
leave -= num[pos];
for (int i = pos+1; i < N; ++i) {
if (sum+num[i] > M || leave+sum < M) {  //剩下的不足时不再进行搜索
return false;
}
if (sum+num[i] == M) {
ans[Count++] = num[i];
return true;
}
if (dfs(sum+num[i], i, leave)) {
ans[Count++] = num[i];
return true;
}
}
return false;
}

2、Codeforces 217A Ice Skating

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <bitset>
#include <ctime>
#include <cctype>

using namespace std;

#define lson low, mid, _id<<1
#define rson mid + 1, high, _id<<1|1

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> Pair;

const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int maxn = 100 + 10;

int n, cnt = 0;
int x[maxn], y[maxn];
bool vis[maxn];

void dfs(int a, int b);

int main() {
#ifdef Floyd
freopen("in.txt", "r", stdin);
#endif
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d %d", &x[i], &y[i]);
}
memset(vis, false, sizeof(vis));
for (int i = 0; i < n; ++i) {
if (!vis[i]) {
++cnt;
dfs(x[i], y[i]);
}
}
printf("%d\n", cnt - 1);
return 0;
}

void dfs(int a, int b) {
for (int i = 0; i < n; ++i) {
if (!vis[i] && (a == x[i] || b == y[i])) {
vis[i] = true;
dfs(x[i], y[i]);
}
}
}

3、HDU 5305 Friends

参考：http://blog.csdn.net/yeguxin/article/details/47042115

解题思路：

dfs枚举边的状态，直接枚举到m条边后再判断会TLE

加判断剪枝后，若枚举的边达到m条即是一种解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <cmath>
#include <cctype>
#include <ctime>
#include <cassert>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define eps 1e-9

typedef long long ll;
typedef pair<int, int> pii;

const int INF = 0x7fffffff;
int T, n, m, x, y, ans = 0;
int key[10], Left[30], Right[30], key_a[10], key_b[10];
bool flag = true;

void dfs(int cnt) {
if (cnt == m) { ++ans; return; }
int l = Left[cnt], r = Right[cnt];
if (key_a[l] < (key[l] / 2) && key_a[r] < (key[r] / 2)) {
++key_a[l]; ++key_a[r]; dfs(cnt + 1); --key_a[l]; --key_a[r];
}
if (key_b[l] < (key[l] / 2) && key_b[r] < (key[r] / 2)) {
++key_b[l]; ++key_b[r]; dfs(cnt + 1); --key_b[l]; --key_b[r];
}
}

int main() {
#ifdef __AiR_H
freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
#endif // __AiR_H
scanf("%d", &T);
while (T--) {
scanf("%d %d", &n, &m);
memset(key, 0, sizeof(key));
REP(i, m) { scanf("%d %d", &x, &y); Left[i] = x; Right[i] = y; ++key[x]; ++key[y]; }
flag = true;
for (int i = 1; i <= n; ++i) {
if (key[i] % 2 != 0) { flag = false; break; }
}
if (!flag) { printf("0\n"); continue; }
memset(key_a, 0, sizeof(key_a)); memset(key_b, 0, sizeof(key_b));
ans = 0; dfs(0); printf("%d\n", ans);
}
#ifdef __AiR_H
printf("Time used = %.2fs\n", (double)clock() / CLOCKS_PER_SEC);
#endif // __AiR_H
return 0;
}