标签:
题意
根据图中公式由A[]构造B[][],现在给你B,问你存不存在一个数组A使之成立。
题解:对于每一位进行2-sat求解。
比赛半个小时时间,没做出来……
一直T。
因为本身对算法不确定,所以也不知道怎么办
赛后才发现是数组开小了、、、、
真坑啊。。。
#include <bits/stdc++.h> using namespace std; const int N = 1005; int a[N], b[N][N]; struct Edge { int from, to, next; } edge[N*N*4]; int head[N], cntE; void addedge(int u, int v) { edge[cntE].from = u; edge[cntE].to = v; edge[cntE].next = head[u]; head[u] = cntE++; } int dfn[N], low[N], idx; int stk[N], top; int in[N]; int kind[N], cnt; void tarjan(int u) { dfn[u] = low[u] = ++idx; in[u] = true; stk[++top] = u; for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].to; if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]); else if (in[v]) low[u] = min(low[u], dfn[v]); } if (low[u] == dfn[u]) { ++cnt; while (1) { int v = stk[top--]; kind[v] = cnt; in[v] = false; if (v == u) break; } } } int opp[N], ind[N], col[N]; bool topsort(int n) { for (int i = 0; i < 2*n; ++i) { if (!dfn[i]) tarjan(i); } for (int i = 0; i < n; ++i) { int k1 = kind[i], k2 = kind[i+n]; if (k1 == k2) return false; } return true; } void init() { cntE = 0; memset(head, -1, sizeof head); memset(dfn, 0, sizeof dfn); memset(in, false, sizeof in); idx = top = cnt = 0; memset(ind, 0, sizeof ind); memset(col, 0, sizeof col); } int main() { //freopen("in.txt", "r", stdin); int n; while (~scanf("%d", &n)) { for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { scanf("%d", &b[i][j]); } } bool fg = true; for (int i = 0; i < n; ++i) { for (int j = i; j < n; ++j) { if (i == j) { if (b[i][j] != 0) { fg = false; break; } } else if (b[i][j] != b[j][i]) { fg = false; break; } } if (!fg) break; } if (!fg) { printf("NO\n"); continue; } for (int w = 0; w <= 30; ++w) { init(); int cnt = 0; for (int i = 0; i < n; ++i) { for (int j = i+1; j < n; ++j) { //if (i == j) continue; if (i % 2 == 0 && j % 2 == 0) { if (b[i][j] & (1 << w)) { addedge(i, i+n); addedge(j, j+n); } else { addedge(j+n, i); addedge(i+n, j); } } else if (i % 2 == 1 && j % 2 == 1) { if (b[i][j] & (1 << w)) { addedge(j, i+n); addedge(i, j+n); } else { addedge(i+n, i); addedge(j+n, j); } } else { if (b[i][j] & (1 << w)) { addedge(i, j+n); addedge(j, i+n); addedge(j+n, i); addedge(i+n, j); } else { addedge(i, j); addedge(j, i); addedge(i+n, j+n); addedge(j+n, i+n); } } } } if (!topsort(n)) { fg = false; break; } } if (fg) printf("YES\n"); else printf("NO\n"); } return 0; }
标签:
原文地址:http://www.cnblogs.com/wenruo/p/5958401.html