标签:void pre 二分 lib queue cst ios mes assert
题意:给定一个矩阵的每行的和和每列的和,以及每个格子的限制,让你求出原矩阵。
析:把行看成X,列看成Y,其实就是二分图,然后每个X到每个Y边一条边,然后加一个超级源点和汇点分别向X和Y连边,这样就形成了一个有源汇有上下界的网络,如果有最大流,那么这个矩阵就存在。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 250 + 10; const int maxm = 1e5 + 10; const int mod = 50007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, -1, 0, 1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } struct Edge{ int from, to, cap, flow; }; struct Dinic{ int n, m, s, t; vector<Edge> edges; vector<int> G[maxn]; int d[maxn]; bool vis[maxn]; int cur[maxn]; void init(int n){ this-> n = n; for(int i = 0; i < n; ++i) G[i].cl; edges.cl; } void addEdge(int from, int to, int cap){ edges.pb((Edge){from, to, cap, 0}); edges.pb((Edge){to, from, 0, 0}); m = edges.sz; G[from].pb(m-2); G[to].pb(m-1); } bool bfs(){ ms(vis, 0); vis[s] = 1; d[s] = 1; queue<int> q; q.push(s); while(!q.empty()){ int x = q.front(); q.pop(); for(int i = 0; i < G[x].sz; ++i){ Edge &e = edges[G[x][i]]; if(!vis[e.to] && e.cap > e.flow){ d[e.to] = d[x] + 1; vis[e.to] = 1; q.push(e.to); } } } return vis[t]; } int dfs(int x, int a){ if(x == t || a == 0) return a; int flow = 0, f; for(int &i = cur[x]; i < G[x].sz; ++i){ Edge &e = edges[G[x][i]]; if(d[e.to] == d[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){ e.flow += f; edges[G[x][i]^1].flow -= f; a -= f; flow += f; if(a == 0) break; } } return flow; } int maxflow(int s, int t){ this->s = s; this-> t = t; int flow = 0; while(bfs()){ ms(cur, 0); flow += dfs(s, INF); } return flow; } }; Dinic dinic; int in[maxn]; int down[maxn][maxn], up[maxn][maxn]; bool ok; void judge(int i, int j, int v){ if(v < down[i][j] || v > up[i][j]) ok = false; } int main(){ int T; cin >> T; while(T--){ scanf("%d %d", &n, &m); int s = 0, t = n + m + 1; int S = t + 1, T = S + 1; FOR(i, 1, S) FOR(j, 1, S){ down[i][j] = -1000; up[i][j] = 1000; } ms(in, 0); dinic.init(T + 2); for(int i = 1; i <= n; ++i){ int x; scanf("%d", &x); in[s] -= x; in[i] += x; dinic.addEdge(s, i, 0); } for(int i = 1; i <= m; ++i){ int x; scanf("%d", &x); in[i+n] -= x; in[t] += x; dinic.addEdge(i+n, t, 0); } int num; scanf("%d", &num); char op[5]; int u, v, c; ok = true; while(num--){ scanf("%d %d %s %d", &u, &v, op, &c); if(!ok) continue; if(u && v){ if(op[0] == ‘=‘){ judge(u, v+n, c); up[u][v+n] = down[u][v+n] = c; } else if(op[0] == ‘<‘) up[u][v+n] = min(up[u][v+n], c - 1); else down[u][v+n] = max(down[u][v+n], c + 1); } else if(u){ for(int j = 1; j <= m; ++j){ if(op[0] == ‘=‘){ judge(u, j+n, c); up[u][j+n] = down[u][j+n] = c; } else if(op[0] == ‘<‘) up[u][j+n] = min(up[u][j+n], c - 1); else down[u][j+n] = max(down[u][j+n], c + 1); } } else if(v){ for(int i = 1; i <= n; ++i){ if(op[0] == ‘=‘){ judge(i, v+n, c); up[i][v+n] = down[i][v+n] = c; } else if(op[0] == ‘<‘) up[i][v+n] = min(up[i][v+n], c - 1); else down[i][v+n] = max(down[i][v+n], c + 1); } } else{ for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j){ if(op[0] == ‘=‘){ judge(i, j+n, c); up[i][j+n] = down[i][j+n] = c; } else if(op[0] == ‘<‘) up[i][j+n] = min(up[i][j+n], c - 1); else down[i][j+n] = max(down[i][j+n], c + 1); } } } int ans = 0; if(!ok){ puts("IMPOSSIBLE"); goto A; } for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j){ if(up[i][j+n] < down[i][j+n]) ok = false; dinic.addEdge(i, j+n, up[i][j+n] - down[i][j+n]); in[i] -= down[i][j+n]; in[j+n] += down[i][j+n]; } dinic.addEdge(t, s, INF); for(int i = 0; i <= t; ++i){ if(in[i] > 0) dinic.addEdge(S, i, in[i]), ans += in[i]; if(in[i] < 0) dinic.addEdge(i, T, -in[i]); } if(!ok || ans != dinic.maxflow(S, T)){ puts("IMPOSSIBLE"); goto A; } for(int i = 1; i <= n; ++i){ int cur = 0; for(int j = 0; dinic.edges[dinic.G[i][j]].to != n+1; ++j, cur = j); for(int j = cur, cnt = 0; cnt < m; ++cnt, ++j) printf("%d%c", dinic.edges[dinic.G[i][j]].flow + down[i][cnt+n+1], " \n"[cnt+1==m]); } A:; if(T) puts(""); } return 0; }
标签:void pre 二分 lib queue cst ios mes assert
原文地址:http://www.cnblogs.com/dwtfukgv/p/7631645.html