LightOJ 1237 Cyber Cafe 费用流

题目：http://www.lightoj.com/volume_showproblem.php?problem=1237

题意：有n个顾客来消费，记录下了他们的进入时间和离开时间，但每个人的进入和离开时间不是一一对应的，而是乱序的。顾客的消费为(T-K)的2次方，T为在店里停留的时间，若消费最高为G，现在假设可以自由对应进入和离开时间，问最少收入和最多收入，时间有可能记错，若记错输出impossible

思路：从进入时间向离开时间连边，注意进入时间要严格小于离开时间，计算好每条边的费用，从源点向进入时间连边，从离开时间向汇点连边，跑一边费用流，求出最小收入。把每条边的费用取负重新建一次图跑费用流，求出最大收入，这是因为取负后求最短路实际是之前的最长路

总结：进入时间要严格小于离开时间，因为这个错了两次

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;const int N = 210;const int INF = 0x3f3f3f3f;struct edge{    int st, to, cost, cap, next;}g[N*N*2];int cnt, head[N];int cas = 0;int arr[N], brr[N];int dis[N], pre[N];bool vis[N];void add_edge(int v, int u, int cost, int cap){    g[cnt].st = v, g[cnt].to = u, g[cnt].cost = cost, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;    g[cnt].st = u, g[cnt].to = v, g[cnt].cost = -cost, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;}void spfa(int s, int t){    memset(dis, 0x3f, sizeof dis);    memset(vis, 0, sizeof vis);    memset(pre, -1, sizeof pre);    queue<int> que;    que.push(s);    dis[s] = 0, vis[s] = true;    while(! que.empty())    {        int v = que.front(); que.pop();        vis[v] = false;        for(int i = head[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(g[i].cap > 0 && dis[u] > dis[v] + g[i].cost)            {                dis[u] = dis[v] + g[i].cost;                pre[u] = i;                if(! vis[u])                    que.push(u), vis[u] = true;            }        }    }}int cost_flow(int s, int t, int flow){    int res = 0;    while(flow > 0)    {        spfa(s, t);        if(dis[t] == INF) return -1;        int d = flow;        for(int i = pre[t]; i != -1; i = pre[g[i].st])            d = min(d, g[i].cap);        flow -= d;        res += d * dis[t];        for(int i = pre[t]; i != -1; i = pre[g[i].st])            g[i].cap -= d, g[i^1].cap += d;    }    return res;}int main(){    int t, n, K, G;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d", &n, &K, &G);        for(int i = 1; i <= n; i++)            scanf("%d", &arr[i]);        for(int i = 1; i <= n; i++)            scanf("%d", &brr[i]);        cnt = 0;        memset(head, -1, sizeof head);        for(int i = 1; i <= n; i++)        {            add_edge(0, i, 0, 1);            add_edge(n + i, n + n + 1, 0, 1);            for(int j = 1; j <= n; j++)            {                if(brr[j] <= arr[i]) continue;                int cost = ((brr[j] - arr[i]) - K) * ((brr[j] - arr[i]) - K);                if(cost > G) cost = G;                add_edge(i, n + j, cost, 1);            }        }        int res1 = cost_flow(0, n + n + 1, n);        cnt = 0;        memset(head, -1, sizeof head);        for(int i = 1; i <= n; i++)        {            add_edge(0, i, 0, 1);            add_edge(n + i, n + n + 1, 0, 1);            for(int j = 1; j <= n; j++)            {                if(brr[j] <= arr[i]) continue;                int cost = ((brr[j] - arr[i]) - K) * ((brr[j] - arr[i]) - K);                if(cost > G) cost = G;                add_edge(i, n + j, -cost, 1);            }        }        int res2 = cost_flow(0, n + n + 1, n);        if(res1 == -1 || res2 == -1) printf("Case %d: impossible\n", ++cas);        else printf("Case %d: %d %d\n", ++cas, res1, -res2);    }    return 0;}