P3980 [NOI2008]志愿者招募 费用流 （人有多大胆地有多大产

感觉费用流比网络流的图更难想到，要更大胆。
首先由于日期是连续的，所以图中的点是横向排列的。

这道题有点绕道走的意思，由于一类志愿者是可以服务于一段时间，那我们给第i天连出去多条边，第一条边是流向i+1点的，容量为inf-a【i】，费用为0，
若有i 到 其他点t，费用为w， 则连一条（i， t+1， inf，w）的边

跑一遍费用流，算出结果

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

/*
        
⊂_ヽ
　 ＼＼ Λ＿Λ  来了老弟
　　 ＼('ㅅ')
　　　 >　⌒ヽ
　　　/ 　 へ＼
　　 /　　/　＼＼
　　 ﾚ　ノ　　 ヽ_つ
　　/　/
　 /　/|
　(　(ヽ
　|　|、＼
　| 丿 ＼ ⌒)
　| |　　) /
'ノ )　　Lﾉ

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const ll mod = 2147483648;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/
            
            const int maxn = 1e3+9;
            int a[maxn];
            struct E{
                int v,val,cost;
                int nxt;
            }edge[maxn*maxn];
            int head[maxn],gtot;
            void addedge(int u,int v,int val,int cost){
                edge[gtot].v = v;
                edge[gtot].val = val;
                edge[gtot].cost = cost;
                edge[gtot].nxt = head[u];
                head[u] = gtot++;

                edge[gtot].v = u;
                edge[gtot].val = 0;
                edge[gtot].cost = -cost;
                edge[gtot].nxt = head[v];
                head[v] = gtot++;
            }

            int dis[maxn],pre[maxn],vis[maxn],path[maxn];
            bool spfa(int s,int t){
                memset(dis,inf, sizeof(dis));
                memset(vis, 0, sizeof(vis));
                memset(pre, -1, sizeof(pre));

                dis[s] = 0; vis[s] = 1;
                queue<int>que;
                que.push(s);

                while(!que.empty()){
                    int u = que.front(); que.pop();
                    vis[u] = 0;
                    for(int i=head[u]; ~i; i=edge[i].nxt){
                        int v = edge[i].v, val = edge[i].val, cost = edge[i].cost;
                        if(val > 0 && dis[v] > dis[u] + cost){
                            dis[v] = dis[u] + cost;
                            pre[v] = u; path[v] = i;
                            if(vis[v] == 0){
                                vis[v] = 1;
                                que.push(v);
                            }
                        }
                    }
                }
                return pre[t] != -1;
            }
            int mcmf(int s,int t){
                int flow = 0, cost = 0;
                while(spfa(s, t)){
                    int f = inf;
                    for(int i=t; i!=s; i=pre[i]){
                        f = min(f, edge[path[i]].val);
                    }
                    flow += f;
                    cost += f * dis[t];
                    for(int i=t; i!=s; i=pre[i]){
                        edge[path[i]].val -= f;
                        edge[path[i]^1].val += f;
                    }
                }
                return cost;
            }
int main(){
            memset(head, -1, sizeof(head));
            int n,m;    
            scanf("%d%d", &n, &m);
            rep(i, 1, n) scanf("%d", &a[i]);
            int s = 0, t = n+2;
            addedge(s, 1, inf, 0);
            addedge(n+1, t, inf, 0);
            rep(i, 1, n) addedge(i, i+1, inf - a[i], 0);
            
            while(m --) {
                int u,v,w;
                scanf("%d%d%d", &u, &v, &w);
                addedge(u, v+1, inf, w);
            }
            printf("%d\n", mcmf(s, t));
            
            return 0;
}

标签：cost,人有,val,int,P3980,edge,gtot,include,多大产
来源： https://www.cnblogs.com/ckxkexing/p/10404094.html