什么破网站，多余空格换行都不能有，还nm不报PE/RE只报WA。
少一行换行也不行，这是什么垃圾文本比较。

March of the Penguins

problem

UVA12125

---

solution

枚举冰块，然后判断被枚举冰块是否符合条件。

将冰块拆成两个点，左点和右点，之间流量是可跳跃次数。

超级源点和左点连边，流量为该冰块上的初始企鹅数。

然后两两冰块枚举能否互跳，分别右点连向左点，流量无穷。

最后连一条最后的冰块左点和超级汇点的边，流量无穷。

---

code

#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 205
#define maxm 40005
#define inf 0x7f7f7f7f
queue < int > q;
struct node { int to, nxt, flow, ori; }E[maxm];
int T, n, cnt, tot, s, t;
double d;
double x[maxn], y[maxn];
int dep[maxn], cur[maxn], head[maxn], num[maxn], lim[maxn];double dis( int i, int j ) { return sqrt( ( x[i] - x[j] ) * ( x[i] - x[j] ) + ( y[i] - y[j] ) * ( y[i] - y[j] ) ); }void addedge( int u, int v, int w ) {E[cnt] = { v, head[u], w, w };head[u] = cnt ++;E[cnt] = { u, head[v], 0, 0 };head[v] = cnt ++;
}bool bfs() {for( int i = 0;i <= t;i ++ ) dep[i] = 0;dep[s] = 1; q.push( s );while( ! q.empty() ) {int u = q.front(); q.pop();for( int i = cur[u] = head[u];~ i;i = E[i].nxt ) {int v = E[i].to; if( ! dep[v] and E[i].flow ) {dep[v] = dep[u] + 1;q.push( v );}}}return dep[t];
}int dfs( int u, int cap ) {if( u == t or ! cap ) return cap;int flow = 0;for( int i = cur[u];~ i;i = E[i].nxt ) {cur[u] = i; int v = E[i].to;if( dep[v] == dep[u] + 1 ) {int w = dfs( v, min( cap, E[i].flow ) );if( ! w ) continue;E[i ^ 1].flow += w;E[i].flow -= w;flow += w;cap -= w;if( ! cap ) break;}}return flow;
}int dinic() {int ans = 0;while( bfs() ) ans += dfs( s, inf );return ans;
}void build( int now ) {memset( head, -1, sizeof( head ) ); cnt = 0;for( int i = 1;i <= n;i ++ ) {addedge( s, i, num[i] );addedge( i, i + n, lim[i] );for( int j = i + 1;j <= n;j ++ )if( dis( i, j ) <= d ) {addedge( i + n, j, inf );addedge( j + n, i, inf );}}addedge( now, t, inf );
}int main() {scanf( "%d", &T );while( T -- ) {memset( head, -1, sizeof( head ) );cnt = tot = 0;scanf( "%d %lf", &n, &d );t = n << 1 | 1;for( int i = 1;i <= n;i ++ ) {scanf( "%lf %lf %d %d", &x[i], &y[i], &num[i], &lim[i] );tot += num[i];}bool flag = 0;for( int i = 1;i <= n;i ++ ) {build( i );if( dinic() == tot ) {if( flag ) printf( " " );printf( "%d", i - 1 );flag = 1;}}if( ! flag ) printf( "-1" ); printf( "\n" );}return 0;
}

矩阵解压 Matrix Decompressing

problem

UVA11082

---

solution

显然可以解出每行每列的和。

然后将行列分成左右点两部分（类似二分图）。

超级源点和行连边，流量为行的和。

列和超级汇点两边，流量为列的和。

然后两两行列点之间流量为 $19$。

当超级源点和超级汇点的边都满流时，$i$ 行 $j$ 列之间的边流过的流量 $+1$ 就是 $a[i,j]$ 的值。

解释：

• 每个数大小在 $[1,20]$，所以不用太大。
• 可能出现零流的状况（因为同是最大流但流法不太一样导致有些边代表的位置值计算出来是 $0$，不符合范围）所以提前 $-1$。

---

code

#include <bits/stdc++.h>
using namespace std;
#define maxn 100
#define inf 0x7f7f7f7f
struct node { int to, nxt, flow; }E[maxn * maxn];
int T, n, m, s, t, cnt;
int r[maxn], c[maxn], head[maxn], cur[maxn], dep[maxn];
int ans[maxn][maxn];queue < int > q;void addedge( int u, int v, int w ) {E[cnt] = { v, head[u], w };head[u] = cnt ++;E[cnt] = { u, head[v], 0 };head[v] = cnt ++;
}bool bfs() {memset( dep, 0, sizeof( dep ) );dep[s] = 1; q.push( s );while( ! q.empty() ) {int u = q.front(); q.pop();for( int i = cur[u] = head[u];~ i;i = E[i].nxt ) {int v = E[i].to;if( ! dep[v] and E[i].flow ) {dep[v] = dep[u] + 1;q.push( v );}}}return dep[t];
}int dfs( int u, int cap ) {if( u == t or ! cap ) return cap;int flow = 0;for( int i = cur[u];~ i;i = E[i].nxt ) {cur[u] = i; int v = E[i].to;if( dep[v] == dep[u] + 1 ) {int w = dfs( v, min( cap, E[i].flow ) );if( ! w ) continue;E[i ^ 1].flow += w;E[i].flow -= w;flow += w;cap -= w;if( ! cap ) break;}}return flow;
}int main() {scanf( "%d", &T );for( int Case = 1;Case <= T;Case ++ ) {scanf( "%d %d", &n, &m );for( int i = 1;i <= n;i ++ ) scanf( "%d", &r[i] );for( int i = 1;i <= m;i ++ ) scanf( "%d", &c[i] );for( int i = n;i;i -- ) r[i] = r[i] - r[i - 1];for( int i = m;i;i -- ) c[i] = c[i] - c[i - 1];cnt = s = 0, t = n + m + 1;memset( head, -1, sizeof( head ) );for( int i = 1;i <= n;i ++ ) addedge( s, i, r[i] - m );for( int i = 1;i <= m;i ++ ) addedge( i + n, t, c[i] - n );for( int i = 1;i <= n;i ++ ) for( int j = 1;j <= m;j ++ ) addedge( i, j + n, 19 );while( bfs() ) dfs( s, inf );for( int i = 1;i <= n;i ++ ) for( int j = head[i];~ j;j = E[j].nxt ) ans[i][E[j].to - n] = 20 - E[j].flow;printf( "Matrix %d\n", Case );for( int i = 1;i <= n;i ++ ) { for( int j = 1;j <= m;j ++ ) printf( "%d ", ans[i][j] ); printf( "\n" ); }}return 0;
}

Buy one, get the rest free

problem

UVA10983

---

solution

观察到总天数 $d\le 10$，非常小，很套路的拆点。

将一个城市拆成 $d$ 个点，表示第 $d$ 天的城市。

一架航班就连对应两座城的两个时间代表点。

一座城市之间也有天数小的与天数大的点连边。

超级源点和每座城市的第 $0$ 点代表点连边，流量为该城市人数。

---

code

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 2005
#define inf 0x7f7f7f7f
struct node { int u, v, c, p, e; }flight[maxn];
struct edge { int to, nxt, flow; }E[maxn << 3];
int T, n, m, d, cnt, tot, s, t;
int head[maxn], dep[maxn], cur[maxn], w[maxn];
queue < int > q;void addedge( int u, int v, int w ) {E[cnt] = { v, head[u], w };head[u] = cnt ++;E[cnt] = { u, head[v], 0 };head[v] = cnt ++;
}bool bfs() {memset( dep, 0, sizeof( dep ) );dep[s] = 1; q.push( s );while( ! q.empty() ) {int u = q.front(); q.pop();for( int i = cur[u] = head[u];~ i;i = E[i].nxt ) {int v = E[i].to;if( ! dep[v] and E[i].flow ) {dep[v] = dep[u] + 1;q.push( v );}}}return dep[( d + 1 ) * n];
}int dfs( int u, int cap ) {if( u == n * ( d + 1 ) or ! cap ) return cap;int flow = 0;for( int i = cur[u];~ i;i = E[i].nxt ) {cur[u] = i; int v = E[i].to;if( dep[v] == dep[u] + 1 ) {int w = dfs( v, min( cap, E[i].flow ) );if( ! w ) continue;E[i ^ 1].flow += w;E[i].flow -= w;flow += w;cap -= w;if( ! cap ) break;}}return flow;
}int dinic() {int ans = 0;while( bfs() ) ans += dfs( s, inf );return ans;
}bool check( int pos ) {memset( head, -1, sizeof( head ) ); cnt = 0;for( int i = 1;i <= n;i ++ ) {addedge( s, i, w[i] );for( int j = 0;j < d;j ++ )addedge( i + j * n, i + ( j + 1 ) * n, inf );}for( int i = 1;i <= pos;i ++ ) {int u = flight[i].u, v = flight[i].v, c = flight[i].c, e = flight[i].e;addedge( u + e * n, v + ( e + 1 ) * n, c );}return dinic() == tot;
}int main() {scanf( "%d", &T );for( int Case = 1;Case <= T;Case ++ ) {scanf( "%d %d %d", &n, &d, &m );for( int i = 1, u, v, c, p, e;i <= m;i ++ ) {scanf( "%d %d %d %d %d", &u, &v, &c, &p, &e );flight[i] = { u, v, c, p, e };}tot = 0;for( int i = 1;i <= n;i ++ ) scanf( "%d", &w[i] ), tot += w[i];sort( flight + 1, flight + m + 1, []( node x, node y ) { return x.p < y.p; } );int l = 1, r = m, ans = -1;while( l <= r ) {int mid = ( l + r ) >> 1;if( check( mid ) ) ans = flight[mid].p, r = mid - 1;else l = mid + 1;}if( ~ ans ) printf( "Case #%d: %d\n", Case, ans );else printf( "Case #%d: Impossible\n", Case );}return 0;
}

The K-League

problem

UVA1306

---

solution

枚举每个队伍成为最后的冠军，然后钦定其剩余的比赛肯定都是这个队伍赢。

然后就是剩下队伍之间的比赛。

比赛建立 $n\ast n$ 个点（两个队伍之间的比赛)，队伍 $n$ 个点。

超级源点和比赛连边，流量为两个队伍之间要进行的比赛场数。

比赛和比赛的两个队伍连边，流量为无穷。

队伍和超级汇点连边，流量为冠军队伍赢的场数减去本队伍已经赢的场数（保证所有队伍赢的场数不能超过冠军队伍）。

当超级源点的边都满流时，证明合法。

---

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 700
#define inf 0x7f7f7f7f
int T, n, cnt, s, t;
int w[maxn], d[maxn], head[maxn], dep[maxn], cur[maxn];
int a[maxn][maxn];
struct node { int to, nxt, flow; }E[maxn * maxn];
queue < int > q;void addedge( int u, int v, int w ) {E[cnt] = { v, head[u], w };head[u] = cnt ++;E[cnt] = { u, head[v], 0 };head[v] = cnt ++;
}int build( int x ) {memset( head, -1, sizeof( head ) ); cnt = 0;int tot = w[x];for( int i = 1;i <= n;i ++ ) tot += a[x][i];for( int i = 1;i <= n;i ++ ) if( w[i] > tot ) return -1;for( int i = 1;i <= n;i ++ )for( int j = i + 1;j <= n;j ++ )if( i ^ x and j ^ x and a[i][j] ) {addedge( s, ( i - 1 ) * n + j, a[i][j] );addedge( ( i - 1 ) * n + j, i + n * n, inf );addedge( ( i - 1 ) * n + j, j + n * n, inf );}for( int i = 1;i <= n;i ++ ) if( i ^ x ) addedge( i + n * n, t, tot - w[i] );return tot - w[x];
}bool bfs() {memset( dep, 0, sizeof( dep ) );dep[s] = 1; q.push( s );while( ! q.empty() ) {int u = q.front(); q.pop();for( int i = cur[u] = head[u];~ i;i = E[i].nxt ) {int v = E[i].to;if( ! dep[v] and E[i].flow ) {dep[v] = dep[u] + 1;q.push( v );}}}return dep[t];
}int dfs( int u, int cap ) {if( u == t or ! cap ) return cap;int flow = 0;for( int i = cur[u];~ i;i = E[i].nxt ) {cur[u] = i; int v = E[i].to;if( dep[v] == dep[u] + 1 ) {int w = dfs( v, min( cap, E[i].flow ) );if( ! w ) continue;E[i ^ 1].flow += w;E[i].flow -= w;flow += w;cap -= w;if( ! cap ) break;}}return flow;
}int dinic() {int ans = 0;while( bfs() ) ans += dfs( s, inf );return ans;
}int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d", &n ); s = 0, t = n * ( n + 1 ) + 1;for( int i = 1;i <= n;i ++ ) scanf( "%d %d", &w[i], &d[i] );int tot = 0;for( int i = 1;i <= n;i ++ ) for( int j = 1;j <= n;j ++ ) scanf( "%d", &a[i][j] ), tot += a[i][j];tot >>= 1;for( int i = 1, flag = 0;i <= n;i ++ ) {int win = build( i );if( win == -1 ) continue;if( dinic() + win == tot ) {if( flag ) printf( " %d", i );else printf( "%d", i );flag = 1;}}printf( "\n" );}return 0;
}

混合图的欧拉回路 Euler Circuit

problem

UVA10735

---

solution

欧拉回路的条件就是出边数等于入边数。

不妨设 $d_i:i$ 点的出边减去入边的数量。

当所有的 $d$ 都为 $0$ 的时候，证明是一个欧拉回路。

给所有无向边随意钦定一个方向。

改变一条边的顺序就是一个点度数 $-1/+1$ 的区别。

可以看作是流量 $1$ 的流通。

把 $d>0$ 的归入左边点，$d<0$ 的归入右边点。

超级源点和左边点连边，流量为左边点多出的数量。

右边点和超级汇点连边，流量为右边点少掉的数量。

无向边就成为左边点和右边点之间流量为 $1$ 的边。

当超级源点超级汇点都满流时，证明欧拉回路出现。

最后根据左右点之间边流量为 $1/0$ 判断方向是否改变，输出这个欧拉回路即可。

---

code

#include <bits/stdc++.h>
using namespace std;
#define maxn 2000
#define inf 0x7f7f7f7f
struct node { int to, nxt, flow; }E[maxn];
pair < int, int > edge[maxn];
vector < int > G[maxn];
queue < int > q;
int n, m, cnt, S, T, sum;
int head[maxn], cur[maxn], dep[maxn], d[maxn], id[maxn];void addedge( int u, int v, int w ) {E[cnt] = { v, head[u], w };head[u] = cnt ++;E[cnt] = { u, head[v], 0 };head[v] = cnt ++;
}bool bfs() {memset( dep, 0, sizeof( dep ) );dep[S] = 1; q.push( S );while( ! q.empty() ) {int u = q.front(); q.pop();for( int i = cur[u] = head[u];~ i;i = E[i].nxt ) {int v = E[i].to;if( ! dep[v] and E[i].flow ) {dep[v] = dep[u] + 1;q.push( v );}}}return dep[T];
}int dfs( int u, int cap ) {if( u == T or ! cap ) return cap;int flow = 0;for( int i = cur[u];~ i;i = E[i].nxt ) {cur[u] = i; int v = E[i].to;if( dep[v] == dep[u] + 1 ) {int w = dfs( v, min( cap, E[i].flow ) );if( ! w ) continue;E[i ^ 1].flow += w;E[i].flow -= w;flow += w;cap -= w;if( ! cap ) break;}}return flow;
}int dinic() {int ans = 0;while( bfs() ) ans += dfs( S, inf );return ans;
}void dfs( int u ) {while( ! G[u].empty() ) {int v = G[u].back();G[u].pop_back();dfs( v );printf( " %d", u );}
}bool init() {scanf( "%d %d", &n, &m );cnt = sum = S = 0, T = n + 1;memset( head, -1, sizeof( head ) );memset( d, 0, sizeof( d ) );for( int i = 1;i <= n;i ++ ) G[i].clear();for( int i = 1, u, v;i <= m;i ++ ) {char ch[5];scanf( "%d %d %S", &u, &v, ch );edge[i] = { u, v };d[u] ++, d[v] --;if( ch[0] == 'U' ) id[i] = cnt, addedge( u, v, 1 );else id[i] = -1;}for( int i = 1;i <= n;i ++ ) {if( d[i] & 1 ) return 0;d[i] >>= 1;if( d[i] > 0 ) sum += d[i], addedge( S, i, d[i] );else addedge( i, T, -d[i] );}return 1;
}void solve() {if( dinic() != sum ) { printf( "No euler circuit exist\n" ); return; }#define u edge[i].first#define v edge[i].secondfor( int i = 1;i <= m;i ++ )if( id[i] >= 0 and ! E[id[i]].flow ) G[u].push_back( v );else G[v].push_back( u );#undef u#undef vprintf( "1" );dfs( 1 );puts( "" );
}int main() {int T;scanf( "%d", &T );while( T -- ) {if( ! init() ) printf( "No euler circuit exist\n" );else solve();if( T ) printf( "\n" );}return 0;
}