【洛谷】P3980 [NOI2008]志愿者招募

我居然现在才会用费用流解线性规划……

当然这里解决的一类问题比较特殊

以式子作为点，变量作为边，然后要求就是变量在不同的式子里出现了两次，系数一次为+1，一次为-1

这样的话就作为了一个出度和一个入度，和边正好对应了

我们设每种志愿者选择人数是\(x_{i}\)

我们的限制是

\[\left\{\begin{matrix}
x_{i} \geq 0\\
x_{1} + x_{2} \geq A_{1}\\
x_{1} + x_{3} \geq A_{2}\\
x_{3} + x_{4} \geq A_{3}
\end{matrix}\right.
\]

假如数据是这样的吧～

但是我们需要流量平衡，于是我们新加几个变量

\[\left\{\begin{matrix}
x_{i} \geq 0\\
y_{i} \geq 0\\
x_{1} + x_{2} - y_{1} - A_{1} = 0\\
x_{1} + x_{3} - y_{2} - A_{2} = 0\\
x_{3} + x_{4} - y_{3} - A_{3} = 0
\end{matrix}\right.
\]

然后呢我们在前面和后面补上0 = 0

差分后就是

\[\left\{\begin{matrix}
x_{i} \geq 0\\
y_{i} \geq 0\\
x_{1} + x_{2} - y_{1} - A_{1} = 0\\
x_{1} + x_{3} - y_{2} - A_{2} = 0\\
x_{3} + x_{4} - y_{3} - A_{3} = 0
\end{matrix}\right.
\]

\[\left\{\begin{matrix}
x_{i} \geq 0\\
y_{i} \geq 0\\
x_{1} + x_{2} - y_{1} - A_{1} = 0\\
-x_{2} + x_{3} - y_{2} + y_{1} + A_{1} - A_{2} = 0\\
-x_{1} + x_{4} - y_{3} + y_{2} + A_{2} - A_{3} = 0 \\
-x_{3} - x_{4} + y_{3} + A_{3} = 0
\end{matrix}\right.
\]

这样的话就满足我们的限制了

我们对于每个式子建一个点，我们把\(-x_{1}\)的式子向\(x_{1}\)连一条边，容量为正无穷，费用为志愿者的费用

\(y\)同理，不过费用是0

然后再看常数项，如果常数项是负的，则向汇点流一条容量为常数项绝对值的边，如果是正的，则源点向这个式子流一条容量为常数项的边

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define eps 1e-10

#define ba 47

#define MAXN 200005

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    while(c < '0' || c > '9') {

    if(c == '-') f = -1;

    c = getchar();

    }

    while(c >= '0' && c <= '9') {

    res = res * 10 +c - '0';

    c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) {

    out(x / 10);

    }

    putchar('0' + x % 10);

}

struct node {

    int to,next,cap;int64 val;

}E[1000005];

int head[1005],sumE = 1,Ncnt,S,T,cur[1005];

int64 ans;

int N,M;

int st[10005],ed[10005],cs[10005],a[1005];

bool vis[1005];

void add(int u,int v,int c,int a) {

    E[++sumE].to = v;

    E[sumE].next = head[u];

    E[sumE].cap = c;

    E[sumE].val = a;

    head[u] = sumE;

}

void addtwo(int u,int v,int c,int a) {

    add(u,v,c,a);add(v,u,0,-a);

}

int64 dis[1005];

bool inq[1005];

queue<int> Q;

bool SPFA() {

    for(int i = 1 ; i <= Ncnt ; ++i) dis[i] = 1e18;

    dis[S] = 0;Q.push(S);

    memset(inq,0,sizeof(inq));

    while(!Q.empty()) {

    int u = Q.front();Q.pop();

    inq[u] = 0;

    for(int i = head[u] ; i ; i = E[i].next) {

        int v = E[i].to;

        if(E[i].cap) {

        if(dis[u] + E[i].val < dis[v]) {

            dis[v] = dis[u] + E[i].val;

            if(!inq[v]) Q.push(v);

        }

        }

    }

    }

    return dis[T] < 1e18;

}

int dfs(int u,int aug) {

    if(u == T) {

    ans += aug * dis[T];

    return aug;

    }

    vis[u] = 1;

    for(int i = head[u] ; i ; i = E[i].next) {

    int v = E[i].to;

    if(!vis[v] && E[i].cap && dis[v] == dis[u] + E[i].val) {

        int t = dfs(v,min(E[i].cap,aug));

        if(t) {

        E[i].cap -= t;

        E[i ^ 1].cap += t;

        return t;

        }

    }

    }

    return 0;

}

void MCMF() {

    while(SPFA()) {

    do {

        memset(vis,0,sizeof(vis));

    }while(dfs(S,0x7fffffff));

    }

}

void Solve() {

    read(N);read(M);

    S = N + 2;T = N + 3;Ncnt = N + 3;

    for(int i = 1 ; i <= N ; ++i) {

    read(a[i]);

    if(a[i] > a[i - 1]) addtwo(i,T,a[i] - a[i - 1],0);

    else addtwo(S,i,a[i - 1] - a[i],0);

    addtwo(i,i + 1,0x7fffffff,0);

    }

    addtwo(S,N + 1,a[N],0);

    for(int i = 1 ; i <= M ; ++i) {

    read(st[i]);read(ed[i]);read(cs[i]);

    addtwo(ed[i] + 1,st[i],0x7fffffff,cs[i]);

    }

    MCMF();

    out(ans);enter;

}

int main(){

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

    return 0;

}