費用流

費用流模板 LOJ

LOJ上模板68ms還挺快

要跑一遍最大費用最大流和最小費用最大流每次跑之前重新建圖範圍很小沒什麼問題

每次SPFA找出最短/最長路然後blabla 把費用W當成邊權

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 5007;
#define Min(_A,_B) (_A>_B?_B:_A)
const int inf = 0x3f3f3f3f;
struct edge {
	int v, c, w, nxt;
}e[maxn << 1];

int head[maxn], eid = 0, d[maxn], pre[maxn], s, t, a[101], b[101], n, m, h[101][101];
bool inq[maxn];
//帶權二分圖匹配
void insert(int u, int v, int c, int w) {
	e[eid].v = v;  e[eid].w = w; e[eid].c = c;e[eid].nxt = head[u];  head[u] = eid++;
	e[eid].v = u;  e[eid].nxt = head[v];  e[eid].w = -w; e[eid].c = 0;  head[v] = eid++;
}

bool spfa() {
	memset(inq, 0, sizeof(inq));
	memset(d, -inf, sizeof(d));
	memset(pre, -1, sizeof(pre));
	queue<int> q;
	q.push(s);
	inq[s] = true;
	d[s] = 0;
	int u, v, i;
	while (!q.empty()) {
		u = q.front(); q.pop();
		inq[u] = false;
		for (i = head[u]; ~i; i = e[i].nxt) {
			if (e[i].c) {
				v = e[i].v;
				if (d[u] + e[i].w>d[v]) {//最長路解決最大費用最大流問題
					d[v] = d[u] + e[i].w;
					pre[v] = i;
					if (!inq[v]) {
						inq[v] = true;
						q.push(v);
					}
				}
			}
		}
	}
	return pre[t] != -1;
}
bool spfa_min() {
	memset(inq, 0, sizeof(inq));
	memset(d, inf, sizeof(d));
	memset(pre, -1, sizeof(pre));
	queue<int> q;
	q.push(s);
	inq[s] = true;
	d[s] = 0;
	int u, v, i;
	while (!q.empty()) {
		u = q.front(); q.pop();
		inq[u] = false;
		for (i = head[u]; ~i; i = e[i].nxt) {
			if (e[i].c) {
				v = e[i].v;
				if (d[u] + e[i].w<d[v]) {//最短路解決最小費用最大流問題
					d[v] = d[u] + e[i].w;
					pre[v] = i;
					if (!inq[v]) {
						inq[v] = true;
						q.push(v);
					}
				}
			}
		}
	}
	return pre[t] != -1;
}

int costflow() {
	int res = 0;
	while (spfa()) {
		int i, flow = inf;
		for (i = t; i != s; i = e[pre[i] ^ 1].v) {
			flow = Min(flow, e[pre[i]].c);//尋找路徑上的最小流量
		}
		for (i = t; i != s; i = e[pre[i] ^ 1].v) {
			e[pre[i] ^ 1].c += flow;
			e[pre[i]].c -= flow;
			res += e[pre[i]].w*flow;
		}
	}
	return res;
}
int costflow_min() {
	int res = 0;
	while (spfa_min()) {
		int i, flow = inf;
		for (i = t; i != s; i = e[pre[i] ^ 1].v) {
			flow = Min(flow, e[pre[i]].c);
		}
		for (i = t; i != s; i = e[pre[i] ^ 1].v) {
			e[pre[i] ^ 1].c += flow;
			e[pre[i]].c -= flow;
			res += e[pre[i]].w*flow;
		}
	}
	return res;
}
inline int read() {
	int s = 0, f = 1; char c = getchar(); while (c<'0' || c>'9') { if (c == '-') f = -1; c = getchar(); }
	while (c >= '0'&&c <= '9') { s = s * 10 + c - '0'; c = getchar(); }
	return s*f;
}

int main() {
	m = read(); n = read(); int i, j;
	memset(head, -1, sizeof(head));
	s = 0; t = n+m+1;
	for (i = 1; i <= m; i++) {
		a[i] = read();
		insert(s, i, a[i], 0);
	}
	for (i = 1; i <= n; i++) {
		b[i] = read();
		insert(i + m, t, b[i], 0);
	}

	for (i = 1; i <= m; i++)
		for (j = 1; j <= n; j++) {
			h[i][j] = read();
			insert(i, j + m, inf, h[i][j]);
	}
	printf("%d\n", costflow_min());
	eid = 0;
	memset(head, -1, sizeof(head));
	for (i = 1; i <= m; i++) insert(s, i, a[i], 0);
	for (i = 1; i <= n; i++) insert(i + m, t, b[i], 0);
	for (i = 1; i <= m; i++)
		for (j = 1; j <= n; j++)
			insert(i, j + m, inf, h[i][j]);
	printf("%d", costflow());
	//getchar();
	return 0;
}

SCOI2007 修車

拆點把每個師傅拆成n個點表示倒數第i個修的是xxx

比如師傅i按順序修了三輛車，時間分別是T1，T2,T3,那麼總時間是T1+（T1+T2） +（T1+T2+T3）

那麼可以設倒數第i個修的車對應的弧費用爲T*i

BZOJ上TLE了時限1s過不去但是洛谷AC了

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define Max(_A,_B) (_A>_B?_A:_B)
#define Min(_A,_B) (_A>_B?_B:_A)
const int inf = 0x3f3f3f3f;
using namespace std;
const int maxn = 370007;//10*60*60
int n, m;
struct edge {
	int v, w, c, nxt;
}e[maxn<<1];
int p[maxn], eid = 0, dis[maxn],pre[maxn],s,t;
bool inq[maxn];
void insert(int u, int v, int c, int w) {
	e[eid].v = v; 
	e[eid].w = w;
	e[eid].c = c;
	e[eid].nxt = p[u];
	p[u] = eid++;
}
void addedge(int u, int v, int c, int w) {
	insert(u, v, c, w);
	insert(v, u, 0, -w);
}
bool spfa() {
	queue<int> q;
	q.push(s);
	memset(dis, inf, sizeof(dis));
	memset(pre, -1, sizeof(pre));
	memset(inq, false, sizeof(inq));
	dis[s] = 0;
	inq[s] = true;
	while (!q.empty()) {
		int u = q.front(); 
		q.pop();
		inq[u] = false;
		for (int i = p[u]; i!=-1; i = e[i].nxt){
			if (e[i].c) {
				int v = e[i].v;
				if (dis[u] + e[i].w < dis[v]) {
					dis[v] = dis[u] + e[i].w;
					pre[v] = i;
					if (!inq[v]) {
						inq[v] = true;
						q.push(v);
					}
				}
			}
		}
	}
	return pre[t] != -1;
}
double cost_flow() {
	double res = 0;
	while (spfa()) {
		int flow = inf;
		for (int i = t; i != s; i = e[pre[i] ^ 1].v) {
			flow = Min(flow, e[pre[i]].c);
			//printf("1");
		}
		for (int i = t; i != s; i = e[pre[i] ^ 1].v) {
			e[pre[i]].c -= flow;
			e[pre[i] ^ 1].c += flow;
			res += e[pre[i]].w*flow;
		}
	}

	return res;
}
inline int read() {
	int s = 0, f = 1; char c = getchar(); while (c<'0' || c>'9') { if (c == '-') f = -1; c = getchar(); }
	while (c >= '0'&&c <= '9') { s = s * 10 + c - '0'; c = getchar(); }
	return s*f;
}
int g[61][11];
int main() {
	m = read();
	n = read();
	s = 0; t = m*n+n+1;
	memset(p, -1, sizeof(p));
	for(int i=1;i<=n;i++)
		for (int j = 1; j <= m; j++) {
			g[i][j] = read();
		}
	for (int i = 1; i <= n*m; i++) addedge(s, i, 1, 0);
	for (int i = n*m + 1; i <= n*m + n; i++)
		addedge(i, t, 1, 0);
	for (int i = 1; i <= m; i++)
		for (int j = 1; j <= n; j++)//次序
			for (int k = 1; k <= n; k++)//客人編號
				addedge((i - 1)*n + j, m*n + k, 1, g[k][i] * j);
	printf("%.2lf\n", cost_flow()/n );
	//getchar();
	return 0;
}

NOI2012 美食節

建圖方法和SCOI2007修車一樣但是數據更強了需要動態連邊
然後注意到可以不把每道菜拆成P[i]個點而是把S點連向表示該到菜的邊的容量設成P[i] 就可以沿這條弧增廣P[i]次了
在費用流過程中計算邊對應的廚師和菜的次序時取模炸了被m=1的數據卡掉了發現原來的取模方式無法算出廚師做的第$\sum P$道菜要特判調了好久/_ \
動態連邊的原理是，如果最小費用最大流增廣路經過了廚師i做的倒數第j道菜的點之前就一定經過了倒數第j-1道菜

因爲倒數第1道菜的對應弧容量最小（按修車那題的建模方式）

時限有點緊還可以在spfa的同時預處理出對應增廣路的流量記爲incf

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define Min(_A,_B) (_A<_B?_A:_B)
const int maxn = 81007,inf=0x3f3f3f3f;
struct edge {
	int v, w, c, nxt;
}e[500007];
int head[maxn], eid = 0, d[maxn],s,
t,n,m,p[maxn],u,v,pre[maxn],incf[maxn],sp;
bool inq[maxn];
int g[41][101];
void insert(int u, int v, int c, int w) {
	e[eid].v = v; 
	e[eid].c = c; 
	e[eid].w = w; 
	e[eid].nxt = head[u];
	head[u]=eid++;
}
void addedge(int u, int v, int c, int w) {
	insert(u, v, c, w);
	insert(v, u, 0, -w);
}
void init() {
	memset(head, -1, sizeof(head));
	eid = 0;
}
/*動態加邊的思路是 如果增廣路流經x->t 與x相連的菜
表示廚師j做的倒數第x-n-(j-1)*p道菜
根據此題的構圖
	廚師i做的第j道菜耗時g[j][i]
	第j+1道菜耗時g[j+1][i]+g[j][i]
	第j+2道菜耗時g[j+2][i]+g[j+1][i]+g[j][i]
	總時長爲g[j][i]*3+g[j+1][i]*2+g[j+2][i]
	故我們設各個菜與S相連 容量爲pi
	廚師與t相連 每個廚師j佔的點編號範圍爲
	n+p*(j-1)+1到n+p*j
	即把每個廚師拆成至多p個點。
有T[x]<T[x+1] 如果沒有連向x 就不加連向x+1的邊
*/
bool spfa() {
	memset(d, inf, sizeof(d));
	memset(inq, false, sizeof(inq));
	memset(pre, -1, sizeof(pre));
	queue<int> q;q.push(s);
	d[s] = 0;inq[s] = true;incf[s] = inf;
	while (!q.empty()) {
		u = q.front(); q.pop();inq[u] = false;
		for (int i = head[u]; ~i; i = e[i].nxt) {
				if (e[i].c>0&&d[e[i].v] > d[u] + e[i].w) {
					v=e[i].v;
					d[v] = d[u] + e[i].w;pre[v] = i;
					incf[v] = Min(incf[u], e[i].c);
					if (!inq[v]) {	inq[v] = true;q.push(v);}
				}
		}
	}
	return pre[t]!=-1;
}
int ans = 0,x;
void cost_flow() {
	while(spfa()){
		for(int i=t;i!=s;i=e[pre[i]^1].v){
			e[pre[i]].c-=incf[t];
			e[pre[i]^1].c+=incf[t];
		}
		ans+=d[t]*incf[t];
		x=e[pre[t]^1].v;//增廣路連向t的邊（廚師拆出的點）
		//獲取廚師編號j：由x=n+(j-1)*p+k知，
		//j=(x-n-1)/p+1
		//廚師現在是倒數第k道菜
		//k=(x-n-1)%p+1
		int a=(x-n)/sp+1,b=(x-n+1)%sp;
		if(!b) b=sp;
		//printf("%d %d %d\n",x,a,b);
		addedge(x+1,t,1,0);
		for(int i=1;i<=n;i++){
			addedge(i,x+1,1,g[i][a]*b);
		}

	}
}
inline int read(){
	int s=0,f=1;char c=getchar();while(c<'0'||c>'9') {if(c=='-') f=-1;c=getchar();}
	while(c>='0'&&c<='9') {s=s*10+c-'0';c=getchar();}
	return s*f;
}
int main(){
	n=read();m=read();
	s=0;
	init();
	for(int i=1;i<=n;i++){
		p[i]=read();sp+=p[i];
		addedge(s,i,p[i],0);//p[i]道菜匹配p[i]次
	}
	t=sp*m+n+1;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++){	
			g[i][j]=read();
			//連接每個廚師做的倒數第一道菜
		}
	for(int j=1;j<=m;j++){
		x=n+(j-1)*sp+1;//廚師j做的倒數第1道菜與x連
		for(int i=1;i<=n;i++){
			addedge(i,x,1,g[i][j]);
		}
		addedge(x,t,1,0);
	}
	//printf("%d\n",sp);
	cost_flow();
	printf("%d",ans);
	return 0;
}

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