演算法很簡單 spfa+增廣路
1、首先按照邊的費用spfa最短路（如果是求最大費用可用最長路）
2、接著增光路的可行流的調節
3、重複1 2 直到最短路無法到達匯點為止
挺簡單的注意細節即可
poj 3422
這題的建圖很精妙 贊
/** this code is made by linmeichen
* problem:
* type of problem:
* thinking:
* feeling:
*/#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long lld;
typedef unsigned int ud;
#define inf_max 0x3f3f3f3f
#define eatline() char chch;while((chch=getchar())!='\n')continue;
#define memsetmax(a) memset(a,0x3f,sizeof a)
#define memsetzero(a) memset(a,0,sizeof a)
#define memsetmin(a) memset(a,-1,sizeof a)
#define memsetfalse(a) memsetzero(a)
#define pq priority_queue
#define q queue
#define maxn 5010
#define maxm 100008
struct edge
e[maxm];
int cnt;
int head[maxn];
int map[55][55];
int dis[maxn], mark[maxn], pre[maxn];
bool spfa(int s, int t)}}
}return dis[t] != inf_max;
}int argument(int s, int t)
for (int i = pre[t]; i!= -1; i = pre[e[i ^ 1].v])
return maxflow;
}int maxcostflow(int s, int t)
void add_edge(int u, int v, int f, int c)
int main()
for (int i = 1; i <= n;i++)
for (int j = 1; j < n; j++)
for (int i = 1; i < n;i++)
for (int j = 1; j <= n; j++)
add_edge(s, 0, k, 0);
add_edge(2*n*n-1, t, k, 0);
printf("%d\n", -maxcostflow(s, t));
}return 0;
}
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