标签:style io ar color os sp for strong on
题意: 给你一个N ,M 构造一个N*M的矩阵,矩阵中每个元素为0-K;
给你每行的和与每列的和。
如果解法唯一 ,输出解法
如果解法不唯一,输出一句话,
如果没有解法,输出一句话。
题解: 经典建图
s ---> 每个行节点,流量为行和
每个列节点----〉t,流量为列和
每行每列单独连接,流量为K
代码:
#include<stdio.h> #include<queue> #include<iostream> #define INF 1000000000 #define N 1005 using namespace std; int list[N], listt[N], deep[N], tot, n, m, k, flagg, mark_dfs[N]; struct Node { int date, value, next; }cun[2000005]; struct Edge { int x, t; }old, xin; void add(int a, int b, int c) { cun[++tot].date = b; cun[tot].value = c; cun[tot].next = list[a]; list[a] = tot; cun[++tot].date = a; cun[tot].value = 0; cun[tot].next = list[b]; list[b] = tot; } int bfs(int s, int t, int n) { queue<Edge> p; xin.x = s; xin.t = 0; p.push(xin); memset(deep,255,sizeof(deep)); deep[s] = 0; while(!p.empty()) { old = p.front(); p.pop(); for(int i = list[old.x]; i; i = cun[i].next) { int date = cun[i].date; int value = cun[i].value; xin.x = date; xin.t = old.t + 1; if(value == 0 || deep[date] != -1) continue; deep[date] = xin.t; p.push(xin); } } for(int i = 0; i <= n; i++) { listt[i] = list[i]; } return deep[t] != -1; } int minn( int a, int b) { if(a < b) return a; return b; } int dfs(int s, int t, int min) { if(s == t) return min; int neww = 0; for(int i = listt[s]; i; i = cun[i].next) { listt[s] = i; int date = cun[i].date; int value = cun[i].value; if(value == 0 || deep[date] != deep[s] + 1) { continue; } int m = dfs(date, t, minn(value, min - neww)); neww += m; cun[i].value -= m; cun[i^1].value += m; if(neww == min) break; } if(neww == 0) deep[s] = 0; return neww; } int dinic( int s, int t, int n) { int num = 0; while(bfs(s, t, n)) { num += dfs(s, t, INF); } return num; } int dfs_1(int s, int p) { for(int i = list[s]; i; i = cun[i].next) { int date = cun[i].date; if(i == (p^1)) continue; if(cun[i].value) { if(mark_dfs[date]) {return 1;} mark_dfs[date] = 1; if(dfs_1(date, i)) return 1; mark_dfs[date] = 0; } } return 0; } void build() { int a, s = n + m + 10, t = s + 1, sum1 = 0, sum2 = 0; memset(list,0,sizeof(list)); memset(mark_dfs,0,sizeof(mark_dfs)); tot = 1; flagg = 0; for(int i = 1; i <= n; i++ ) { scanf("%d", &a); sum1 += a; add(s, i, a); } for(int i = 1;i <= m; i++ ) { scanf("%d", &a); sum2 += a; add(i + n, t, a); } if(sum1 != sum2) {printf("Impossible\n"); return;} int kk = tot; for(int i = 1; i <= n; i++ ) for(int j = 1; j <= m; j++ ) { add(i, j + n, k); } int mm = dinic ( s, t, t + 10 ), flag = 0; if(mm != sum1) {printf("Impossible\n"); return;} for(int i = 1; i <= n; i++ ) { if( dfs_1(i, -1) ) { flagg = 1; break; } } kk++; if(flagg) printf("Not Unique\n"); else { printf("Unique\n"); for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { if(j == 1) printf("%d",k - cun[kk].value); else printf(" %d",k - cun[kk].value); kk += 2; } printf("\n"); } } } int main() { while(scanf("%d%d%d",&n, &m, &k)!=EOF) { build(); } }
标签:style io ar color os sp for strong on
原文地址:http://blog.csdn.net/q651111937/article/details/41554399