BZOJ1221：

　　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
      
       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
       
        #define MP make_pairtypedef 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\n",ans);    return 0;}
       
      

 

BZOJ2424: [HAOI2010]订货：

    trick：拆点

#include
      
       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
       
        #define MP make_pairtypedef 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<
        
         <
        
       
      

 

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 

   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
      
       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
       
        #define MP make_pairtypedef 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\n",1-pow(2.0,-ans));    }    return 0;}