hdu 5302 Connect the Graph（构造）

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题目链接：hdu 5302 Connect the Graph

当w1或者是b1为奇数时，无解。其他情况下一定可以构造出一条链和多个节点对，只要保证两点之间没有重边就行了。一种间隔1为一对，另一种间隔2为一对。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int N;int jump(int p, int k) {if (p == N)return 2;p += k;if (p > N)return p - 1;return p;}void solve (int n, int a1, int a2, int k) {int p = 1;for (int i = 0; i <= a2; i++) {int t = jump(p, k + 1);printf("%d %d %d\n", p, t, k);p = t;}p = jump(p, k + 1);for (int i = 1; i < a1; i++) {int t = jump(p, k + 1);printf("%d %d %d\n", p, t, k);p = jump(t, k + 1);}}int main () {int cas, w0, w1, w2, b0, b1, b2;scanf("%d", &cas);while (cas--) {scanf("%d%d%d%d%d%d", &w0, &w1, &w2, &b0, &b1, &b2);N = w0 + w1 + w2;if ((w1&1) || (b1&1)) {printf("-1\n");continue;}printf("%d\n", w1 / 2 + b1 / 2 + w2 + b2);solve(N, w1 / 2, w2, 0);solve(N, b1 / 2, b2, 1);}return 0;}

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