POJ 2396 Budget
链接:http://poj.org/problem?id=2396
题意:给定一个M*N的矩阵,给定每行每列的和,以及其中一些值的限定条件,问能否构成一个可行的矩阵。
思路:
添加一个源点,向每行连边,每条边的上下界都为该行的和;添加一个汇点,每列向汇点连边,边的上下界都为该列的和。然后每行向每列连边,边的上下界一开始为(0,INF),之后通过一些限定条件更新。
现在问题成了求一个有源汇有上下界的可行流。只需要再添加一个超级源点,一个超级汇点,并且将原图的汇点向源点连边,求一个可行流即可。
细节:这道题目的构图比较麻烦。
/* ID: [email protected] PROG: LANG: C++ */ #include<map> #include<set> #include<queue> #include<stack> #include<cmath> #include<cstdio> #include<vector> #include<string> #include<fstream> #include<cstring> #include<ctype.h> #include<iostream> #include<algorithm> using namespace std; #define INF (1 << 30) #define LINF (1LL << 60) #define PI acos(-1.0) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i < n; i++) #define per(i, a, n) for (int i = n - 1; i >= a; i--) #define eps 1e-6 #define debug puts("===============") #define pb push_back #define mkp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second #define SZ(x) ((int)(x).size()) #define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m) typedef long long ll; typedef unsigned long long ULL; const int maxn = 2000; const int maxm = 2000000; struct node { int v; // vertex int cap; // capacity int flow; // current flow in this arc int nxt; } e[maxm * 2]; int g[maxn], cnt; int st, ed, n, N, M, S, T; int id[maxm]; void add(int u, int v, int c, int k) { e[++cnt].v = v; e[cnt].cap = c; e[cnt].flow = 0; e[cnt].nxt = g[u]; g[u] = cnt; id[k] = cnt; e[++cnt].v = u; e[cnt].cap = 0; e[cnt].flow = 0; e[cnt].nxt = g[v]; g[v] = cnt; id[0] = 0; } int sum1, sum2; int low[maxm], up[maxm], tot[maxn]; bool flag; bool set_edge(int i, int j, char* str, int c) { int x = (i - 1) * N + j; if (str[0] == '=') { if (low[x] <= c && c <= up[x]) low[x] = up[x] = c; else return false; } else if (str[0] == '<') { if (low[x] < c) up[x] = min(up[x], c - 1); else return false; } else if (str[0] == '>') { if (up[x] > c) low[x] = max(low[x], c + 1); else return false; } return true; } int sum; void init() { memset(g, 0, sizeof(int) * (M + N + 10)); memset(tot, 0, sizeof(int) * (M + N + 10)); cnt = 1; st = M + N + 2, ed = M + N + 3; S = 0, T = M + N + 1; int u, v, c; sum1 = sum2 = 0; for (int i = 1; i <= M; i++) { scanf("%d", &c); tot[S] -= c; tot[i] += c; sum1 += c; } for (int i = 1; i <= N; i++) { scanf("%d", &c); tot[T] += c; tot[i + M] -= c; sum2 += c; } flag = true; if (sum1 != sum2) flag = false; int k; scanf("%d", &k); memset(low, 0, sizeof(int) * (M * N + 10)); fill(up, up + sizeof(int) * (M * N + 10), INF); char str[10]; while(k--) { scanf("%d%d%s%d", &u, &v, str, &c); if (flag) { int rst, red, cst, ced; if (u == 0) rst = 1, red = M; else rst = red = u; if (v == 0) cst = 1, ced = N; else cst = ced = v; for (int i = rst; i <= red; i++) { for (int j = cst; j <= ced; j++) { if (!set_edge(i, j, str, c)) { flag = false; goto FK; } } } FK:; } } if (!flag) return ; add(T, S, INF, 0); for (int i = 1; i <= M; i++) { for (int j = 1; j <= N; j++) { int x = (i - 1) * N + j; tot[i] -= low[x]; tot[j + M] += low[x]; add(i, j + M, up[x] - low[x], x); } } sum = 0; for (int i = 0; i <= M + N + 1; i++) { if (tot[i] > 0) add(st, i, tot[i], 0), sum += tot[i]; if (tot[i] < 0) add(i, ed, -tot[i], 0); } n = ed + 3; } int dist[maxn], numbs[maxn], q[maxn]; void rev_bfs() { int font = 0, rear = 1; for (int i = 0; i <= n; i++) { //n为总点数 dist[i] = maxn; numbs[i] = 0; } q[font] = ed; dist[ed] = 0; numbs[0] = 1; while(font != rear) { int u = q[font++]; for (int i = g[u]; i; i = e[i].nxt) { if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue; dist[e[i].v] = dist[u] + 1; ++numbs[dist[e[i].v]]; q[rear++] = e[i].v; } } } int maxflow() { rev_bfs(); int u, totalflow = 0; int curg[maxn], revpath[maxn]; for(int i = 0; i <= n; ++i) curg[i] = g[i]; u = st; while(dist[st] < n) { if(u == ed) { // find an augmenting path int augflow = INF; for(int i = st; i != ed; i = e[curg[i]].v) augflow = min(augflow, e[curg[i]].cap); for(int i = st; i != ed; i = e[curg[i]].v) { e[curg[i]].cap -= augflow; e[curg[i] ^ 1].cap += augflow; e[curg[i]].flow += augflow; e[curg[i] ^ 1].flow -= augflow; } totalflow += augflow; u = st; } int i; for(i = curg[u]; i; i = e[i].nxt) if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break; if(i) { // find an admissible arc, then Advance curg[u] = i; revpath[e[i].v] = i ^ 1; u = e[i].v; } else { // no admissible arc, then relabel this vertex if(0 == (--numbs[dist[u]])) break; // GAP cut, Important! curg[u] = g[u]; int mindist = n; for(int j = g[u]; j; j = e[j].nxt) if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]); dist[u] = mindist + 1; ++numbs[dist[u]]; if(u != st) u = e[revpath[u]].v; // Backtrack } } return totalflow; } int main () { int T; scanf("%d", &T); while(T--) { scanf("%d%d", &M, &N); init(); if (!flag) puts("IMPOSSIBLE"); else { int dd = maxflow(); if (dd != sum) { puts("IMPOSSIBLE"); continue; } for (int i = 1; i <= M; i++) { for (int j = 1; j <= N; j++) { int x = (i - 1) * N + j; printf("%d%c", low[x] + e[id[x]].flow, j == N ? '\n' : ' '); } } } if (T) puts(""); } return 0; }
时间: 2024-10-14 10:57:31