费用流做题记录

BZOJ1221：http://www.lydsy.com/JudgeOnline/problem.php?id=1221

　　trick：将每天用完的，和要用的分来开处理，避免花费的重叠计算，也就是拆点

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 4e5+10, M = 1e3+20, mod = 1e9+7,inf = 2e7;


int na[N],ans,ans1,S,T,head[N],cnt = 1,n,dis[N],q[N],inq[N],from[N],f,fA,fB;
struct {
    int from,to,next,c,v;
}e[N * 2];
void ins(int u,int v,int w,int c) {
    cnt++;
    e[cnt].from = u;
    e[cnt].to = v;
    e[cnt].v = w;
    e[cnt].c = c;
    e[cnt].next = head[u];
    head[u] = cnt;
}
void insert(int u,int v,int w,int c) {
    ins(u,v,w,c);ins(v,u,0,-c);
}
int spfa() {
    for(int i = 0; i <= T; ++i) dis[i] = inf;
    int t = 0,w = 1;
    dis[S] = q[S] = S;
    inq[S] = 1;
    while(t!=w) {
        int now = q[t++];
        if(t == 200001) t = 0;
        for(int i = head[now]; i; i = e[i].next) {
            if(e[i].v&&dis[e[i].to] > dis[now] + e[i].c)
            {
                from[e[i].to] = i;
                dis[e[i].to] = dis[now] + e[i].c;
                if(!inq[e[i].to]) {
                    inq[e[i].to] = 1;
                    q[w++] = e[i].to;
                    if(w == 200001) w = 0;
                }
            }
        }
        inq[now] = 0;
    }
    if(dis[T] >= inf) return 0;
    return 1;
}
void mcf() {
    int i = from[T],x=inf;
    while(i) {
        x = min(e[i].v,x);
        i = from[e[i].from];
    }
    i = from[T];
    ans1 = x;
    while(i) {
        e[i].v -= x;
        e[i^1].v += x;
        ans += e[i].c*x;
        i = from[e[i].from];
    }
}
int a,b;
int main()
{
    cnt = 1;
    memset(head,0,sizeof(head));
    scanf("%d%d%d%d%d%d",&n,&a,&b,&f,&fA,&fB);
    S = 0,T = 3*n;
    for(int i = 1; i <= n; ++i){
        scanf("%d",&na[i]);
        insert(S,i,inf,0);
        insert(S,i+n,inf,f);
        insert(i+n,T,na[i],0);
        if(i + 1 <= n)
            insert(i,i+1,inf,0);
        if(i + a + 1 <= n)
            insert(i,n+i+a+1,inf,fA);
        if(i + b + 1 <= n)
            insert(i,n+i+b+1,inf,fB);
    }
    ans = 0,ans1 = 0;
    while(spfa()) mcf();
    printf("%d
",ans);
    return 0;
}

BZOJ2424: [HAOI2010]订货：http://www.lydsy.com/JudgeOnline/problem.php?id=2424

    trick：拆点

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 4e5+10, M = 1e3+20, mod = 1e9+7,inf = 2e7;


int ans,ans1,cnt,head[N],from[N],dis[N],inq[N],q[N],S,T;
struct ss{
    int from,to,next,c,v;
}e[N * 2];
void ins(int u,int v,int w,int c) {
    cnt++;
    e[cnt].from = u;
    e[cnt].to = v;
    e[cnt].v = w;
    e[cnt].c = c;
    e[cnt].next = head[u];
    head[u] = cnt;
}
void insert(int u,int v,int w,int c) {
    ins(u,v,w,c);ins(v,u,0,-c);
}
int spfa() {
    for(int i = 0; i <= T; ++i) dis[i] = inf;
    int t = 0,w = 1;
    dis[S] = 0;
    q[t] = S;
    inq[S] = 1;
    while(t != w) {
        int now = q[t++];
        if(t == 200001) t = 0;
        for(int i = head[now]; i; i = e[i].next){
            if(e[i].v && dis[e[i].to] > dis[now] + e[i].c) {
                from[e[i].to] = i;
                dis[e[i].to] = dis[now] + e[i].c;
                if(!inq[e[i].to]) {
                    inq[e[i].to] = 1;
                    q[w++] = e[i].to;
                    if(w == 200001) w = 0;
                }
            }
        }
        inq[now] = 0;
    }
    if(dis[T] >= inf) return 0;
    return 1;
}
void mcf() {
    int i = from[T],x = inf;
    while(i) {
        x = min(e[i].v, x);
        i = from[e[i].from];
    }
    i = from[T];
    ans1 = x;
    while(i) {
      e[i].v -= x;
      e[i^1].v += x;
      ans += e[i].c * x;
      i = from[e[i].from];
    }
}
int n,m,s,x;
int main() {
    cnt = 1;
    memset(head,0,sizeof(head));
    scanf("%d%d%d",&n,&m,&s);
    T = 2*n+1,S = 0;
    for(int i = 1; i <= n; ++i) {
        scanf("%d",&x);
        insert(i+n,T,x,0);
        insert(i,i+n,inf,0);
    }
    for(int i = 1; i <= n; ++i) {
        scanf("%d",&x);
        insert(S,i,inf,x);
        if(i+1<=n)insert(i,i+1,s,m);
    }
    ans1 = 0,ans = 0;
    while(spfa()) mcf();
    cout<<ans<<endl;
    return 0;
}

BZOJ3171: [Tjoi2013]循环格

　trick：每个点的入度 = 出度 = 1，所以，将每个点拆点

　　      S向每个点连容量1费用0的边

　　　　每个点拆出的点向T连容量1，费用0的边

　　　　每个格子向四周连费用0或1的边

HDU5988 http://acm.hdu.edu.cn/showproblem.php?pid=5988

   trick：第一次走这条边的话，我们拆边就行了

　　　　 ans = 1 - 最大不能破坏概率

　　　　不能破坏概率 = (1-p1)*(1-p2)*(……); ————》 2^(log2((1-p1)+log2(1-p2)+log2(……));

　　　　基础费用流

　　　　需要不能破坏概率最大，就将   -( log2((1-p1)+log2(1-p2)+log2(……) ) 求最小， 边费用 就放-log(1-p);就OK

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const double eps = 0.000001;
const int N = 6e5+10, M = 1e3+20, mod = 1e9+7,inf = 2e9;
int cnt,head[N],from[N],inq[N],q[N],S,T;
double ans1,ans,dis[N];
struct ss{
    int from,to,next,v;
    double c;
}e[N * 2];
void ins(int u,int v,int w,double c) {
    cnt++;
    e[cnt].from = u;
    e[cnt].to = v;
    e[cnt].v = w;
    e[cnt].c = c;
    e[cnt].next = head[u];
    head[u] = cnt;
}
void insert(int u,int v,int w,double c) {
    ins(u,v,w,c);ins(v,u,0,-c);
}
int spfa() {
    for(int i = 0; i <= T; ++i) dis[i] = 1.0*inf,from[i] = -1;
    int t = 0,w = 1;
    dis[S] = 0;
    from[S] = 0;
    q[t] = S;
    inq[S] = 1;
    while(t != w) {
        int now = q[t++];
        for(int i = head[now]; i; i = e[i].next){
            if(e[i].v && dis[e[i].to] - dis[now]-e[i].c> eps ) {
                from[e[i].to] = i;
                dis[e[i].to] = dis[now] + e[i].c;
                if(!inq[e[i].to]) {
                    inq[e[i].to] = 1;
                    q[w++] = e[i].to;
                }
            }
        }
        inq[now] = 0;
    }
    if(from[T] == -1) return 0;
    return 1;
}

void mcf() {
    int i = from[T],x = inf;
    while(i) {
        x = min(e[i].v, x);
        i = from[e[i].from];
    }
    i = from[T];
    while(i) {
      e[i].v -= x;
      e[i^1].v += x;
      ans += e[i].c*x;
      i = from[e[i].from];
    }
}
int n,m,x,y,w,Ts;
double p;
int main() {
    scanf("%d",&Ts);
    while(Ts--) {
        scanf("%d%d",&n,&m);
        S = 0,T = n+1;
        cnt = 1;
        memset(head,0,sizeof(head));
        for(int i = 1; i <= n; ++i) {
            scanf("%d%d",&x,&y);
            if(x > y)insert(S,i,x-y,0);
            if(y > x)insert(i,T,y-x,0);
        }
        for(int i = 1; i <= m; ++i) {
            scanf("%d%d%d%lf",&x,&y,&w,&p);
            if(w >= 1) insert(x,y,1,0);
            if(w > 1) insert(x,y,w-1,-log2(1-p));
        }
        ans = 0;
        while(spfa()) mcf();
        printf("%.2f
",1-pow(2.0,-ans));
    }
    return 0;
}