【POJ 2396】 Budget 带上下界网络流 解题报告
来源:互联网 发布:js正则表达式校验邮箱 编辑:程序博客网 时间:2024/06/05 05:47
Budget
Time Limit: 3000MS Memory Limit: 65536K
Special Judge
Description
We are supposed to make a budget proposal for this multi-site competition. The budget proposal is a matrix where the rows represent different kinds of expenses and the columns represent different sites. We had a meeting about this, some time ago where we discussed the sums over different kinds of expenses and sums over different sites. There was also some talk about special constraints: someone mentioned that Computer Center would need at least 2000K Rials for food and someone from Sharif Authorities argued they wouldn’t use more than 30000K Rials for T-shirts. Anyway, we are sure there was more; we will go and try to find some notes from that meeting.
And, by the way, no one really reads budget proposals anyway, so we’ll just have to make sure that it sums up properly and meets all constraints.
Input
The first line of the input contains an integer N, giving the number of test cases. The next line is empty, then, test cases follow: The first line of each test case contains two integers, m and n, giving the number of rows and columns (m <= 200, n <= 20). The second line contains m integers, giving the row sums of the matrix. The third line contains n integers, giving the column sums of the matrix. The fourth line contains an integer c (c < 1000) giving the number of constraints. The next c lines contain the constraints. There is an empty line after each test case.
Each constraint consists of two integers r and q, specifying some entry (or entries) in the matrix (the upper left corner is 1 1 and 0 is interpreted as “ALL”, i.e. 4 0 means all entries on the fourth row and 0 0 means the entire matrix), one element from the set {<, =, >} and one integer v, with the obvious interpretation. For instance, the constraint 1 2 > 5 means that the cell in the 1st row and 2nd column must have an entry strictly greater than 5, and the constraint 4 0 = 3 means that all elements in the fourth row should be equal to 3.
Output
For each case output a matrix of non-negative integers meeting the above constraints or the string “IMPOSSIBLE” if no legal solution exists. Put one empty line between matrices.
Sample Input
2
2 3
8 10
5 6 7
4
0 2 > 2
2 1 = 3
2 3 > 2
2 3 < 5
2 2
4 5
6 7
1
1 1 > 10
Sample Output
2 3 3
3 3 4
IMPOSSIBLE
解题报告
这是一道比较裸的上下界网络流
由题目给出的限制, 建造m + n + 2个节点,
其中m个节点代表行, n个节点代表列
S向每个代表行的节点连接下界l,上节r都为行上数字和的边 记为 [l,r]
同理, 每个代表列的节点向T连边。
对于每个方格(i,j) 看做一条S + i -> S + m + j [low[i][j], up[i][j]]的边进行连接
然后套上上下界网络流的版跑就行了
比如样例1:
比较恶心的是有可能出现0 0 这种对整个图都有限制的输入和非法的输入。。。
下面贴代码 :
#include <cstdio>#include <cstring>#include <queue>typedef long long LL;const int MAXM = 220;const int MAXN = 22;const int INF = 0x3f3f3f3f;template <class T> T min(T a, T b) { return a<b ? a : b;}template <class T> T max(T a, T b) { return a>b ? a : b;}inline int getInt() { int ret = 0; char ch; bool k = false; while((ch = getchar()) < '0' || ch > '9') if(ch == '-') k = true; do {ret *= 10; ret += ch - '0';} while((ch = getchar()) >= '0' && ch <= '9') ; ungetc(ch, stdin); return k ? -ret : ret;}namespace isap { const int MAXD = MAXN * MAXM * 4; const int MAXE = MAXN * MAXM * 200; struct Node { int v,w,bk,nxt; } d[MAXE]; int head[MAXD], etot; inline int addedge(int a, int b, int c) { ++ etot; d[etot].v = b; d[etot].w = c; d[etot].bk = etot + 1; d[etot].nxt = head[a]; head[a] = etot; ++ etot; d[etot].v = a; d[etot].w = 0; d[etot].bk = etot - 1; d[etot].nxt = head[b]; head[b] = etot; return etot; } int dis[MAXD], vd[MAXD]; std :: queue<int> que; void bfs(int s){ memset(dis, 0x3f, sizeof dis); dis[s] = 0; que.push(s); while(!que.empty()) { int u = que.front(); que.pop(); for(int e = head[u]; e; e = d[e].nxt) if(dis[d[e].v] > dis[u] + 1) { dis[d[e].v] = dis[u] + 1; que.push(d[e].v); } } } int S, T, n; int dfs(int u, int val) { if(u == T) return val; int rest = val; for(int e = head[u]; e; e = d[e].nxt) if(d[e].w && dis[d[e].v] + 1 == dis[u]) { int t = dfs(d[e].v, min(d[e].w, rest)); d[e].w -= t; d[d[e].bk].w += t; rest -= t; if(!rest) return val; if(dis[S] == n) return val - rest; } if(rest == val) { -- vd[dis[u]]; if(!vd[dis[u]]) return 0; dis[u] = n; for(int e = head[u]; e; e = d[e].nxt) if(d[e].w) dis[u] = min(dis[u], dis[d[e].v] + 1); ++ vd[dis[u]]; } return val - rest; } int sap (int s, int t, int nn) { S = s; T = t; n = nn; bfs(T); memset(vd, 0, sizeof vd); for(int i = 1; i<=n; i++) if(dis[i] <= n) ++ vd[dis[i]]; int flow = 0; while(dis[S] < n) flow += dfs(S, INF); return flow; } void reset() { etot = 0; memset(head, 0, sizeof head); S = T = n = 0; }}int X, Y;int edge_sum;const int IDMOV = 2;inline int addedgeEx(int a, int b, int c, int d){ //printf("%d to %d : [%d,%d]\n", a,b,c,d); isap :: addedge(X, b, c); isap :: addedge(a, Y, c); edge_sum += c; return isap :: addedge(a, b, d - c) - IDMOV;}int upper[MAXM][MAXN];int lower[MAXM][MAXN];int ans [MAXM][MAXN];int main () { int cas ; scanf("%d", &cas); while(cas -- ){ isap :: reset(); memset(lower, 0xff, sizeof lower); memset(upper, 0x3f, sizeof upper); int n = getInt(), m = getInt(); int S = 1, T = S + n + m + 1; int sum1 = 0, sum2 = 0; X = T + 1, Y = T + 2; int a, b; edge_sum = 0; for(int i = 1; i<=n; i++) { a = getInt(); sum1 += a; addedgeEx(S, S + i, a, a); } for(int i = 1; i<=m; i++) { b = getInt(); sum2 += b; addedgeEx(S + n + i, T, b, b); } int q = getInt(); bool nosol = false; char tmp[10]; while(q --) { a = getInt(); b = getInt(); scanf("%s", tmp); if(tmp[0] == '>') lower[a][b] = max(lower[a][b], getInt()); else if(tmp[0] == '<') upper[a][b] = min(upper[a][b], getInt()); else if(tmp[0] == '='){ int t = getInt(); if(lower[a][b] >= t || upper[a][b] <= t) nosol = true; lower[a][b] = t - 1, upper[a][b] = t + 1; } } if(sum1 != sum2) {puts("IMPOSSIBLE"); if(q)putchar('\n'); continue ;} if(nosol) {puts("IMPOSSIBLE"); if(q)putchar('\n'); continue ;} for(int i = 1; i<=n; i++) for(int j = 1; j<=m; j++){ a = max(max(lower[0][0], lower[i][j]), max(lower[i][0], lower[0][j])) + 1; b = min(min(upper[0][0], upper[i][j]), min(upper[i][0], upper[0][j])) - 1; if(a > b) {nosol = true; break;} else ans[i][j] = addedgeEx(S + i, S + n + j, a, b); } if(nosol) {puts("IMPOSSIBLE"); if(q)putchar('\n'); continue ;} int r = isap :: addedge(T, S, INF); int flow = isap :: sap(X, Y, Y); if(flow != edge_sum) {puts("IMPOSSIBLE"); if(q)putchar('\n'); continue ;} if(isap :: d[r].w != sum1 || isap :: d[r].w != sum2){ puts("IMPOSSIBLE"); if(q)putchar('\n'); } else { for(int i = 1; i<=n; i++){ printf("%d", isap :: d[ans[i][1]].w + isap :: d[ans[i][1] + 2].w); for(int j = 2; j<=m; j++) printf(" %d", isap :: d[ans[i][j]].w + isap :: d[ans[i][j] + 2].w); putchar('\n'); } } }}
- 【POJ 2396】 Budget 带上下界网络流 解题报告
- poj 2396 Budget 带上下界的网络流模型
- POJ 2396 Budget 上下界网络流
- POJ 2396 Budget 上下界网络流
- POJ 2396 Budget 上下界网络流
- poj 2396 Budget 解题报告
- POJ 2396 Budget 有上下界的网络流
- POJ 2396 Budget 上下界网络流 经典题
- poj 2396 Budget 有上下界的网络流
- POJ 2396 Budget 有上下界的网络流
- POJ 2396 Budget(有上下界的网络流)
- poj 2396 Budget (有源汇有上下界的网络流)
- POJ 2396 Budget (上下界网络流)
- POJ 2396 Budget 有上下界的可行网络流
- 带上下界的网络流
- POJ 2396 Budget(无源汇网络有上下界的可行流-Dinic)
- POJ 2396:Budget (有流量上下界的网络流)
- POJ 2396 - Budget 有源汇的上下界可行流
- 第8章 Spring Boot的数据访问
- JavaScript基础系列8---BOM操作
- eclipse git 报 git: 401 Unauthorized 解决办法
- 三国轶事--巴蜀之危 排列组合
- ubuntu常用软件包deb的安装与卸载
- 【POJ 2396】 Budget 带上下界网络流 解题报告
- 让你快速学习python基础笔记002(一起动手实践)
- Redis
- udp_server
- RNN代码解读之char-RNN with TensorFlow(model.py)
- libubox
- 设计模式之模板模式
- Remove Element
- 差分数组的总结