【POJ3680】Intervals 最小费用最大流

题意:给定N个带权开区间，每个区间有一定的权重，现要求从中选一些区间，要求任意点不被超过K个区间覆盖，求最大的总的权重。

思路:做这道题需要对网络流有一定的理解。注意到题目中的约束是每个点不被超过K个区间覆盖，则以这些点为网络流中的点，覆盖次数的限制即流量的限制。想象所有点都在一条数轴上，源点与第一个点相连，汇点与最后一个点相连，容量都为K。每个点一次相连，容量为INF，这样可以保证每个区间都能被选到，对于每个区间，建立容量为1，费用为-Wi的流，因为题目中是开区间，这样建边的话可以保证两个端点之间的流量不超过K,也就是区间间的每个点被选不超过K次。

代码：

struct Edge{    int to,next,cap,flow,cost;  }edge[maxm];int head[maxn],tot;int pre[maxn],dis[maxn];bool vis[maxn];int N;void init(){    tot = 0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int cap,int cost){    edge[tot].to = v;    edge[tot].cap = cap;    edge[tot].cost = cost;    edge[tot].flow = 0;    edge[tot].next = head[u];    head[u] = tot++;    edge[tot].to = u;    edge[tot].cap = 0;    edge[tot].cost = -cost;    edge[tot].flow = 0;    edge[tot].next = head[v];    head[v] = tot++;}int spfa(int s,int t){    queue<int>q;    for(int i = 0;i<N;i++){        dis[i] = inf;        vis[i] = 0;        pre[i] = -1;    }    dis[s] = 0;    vis[s] = 1;    q.push(s);    while(!q.empty()){        int u = q.front();q.pop();        vis[u] = 0;        for(int i = head[u];i!=-1;i = edge[i].next){            int v = edge[i].to;            if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost){                dis[v] = dis[u] + edge[i].cost;                pre[v] = i;                if(!vis[v]){                    vis[v] = 1;                    q.push(v);                }            }        }    }    if(pre[t] == -1) return 0;    return 1;}int MCMF(int s, int t, int &cost) {    int flow = 0;    cost = 0;    while (spfa(s, t)) {        int Min = inf;        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {            if (Min > edge[i].cap - edge[i].flow)                Min = edge[i].cap - edge[i].flow;        }        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]) {            edge[i].flow += Min;            edge[i ^ 1].flow -= Min;            cost += edge[i].cost * Min;        }        flow += Min;    }    return flow;}int a[210],b[210],c[210];vector<int>hs;int main(){    int T,n,k;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&k);        hs.clear();        init();        for(int i = 0;i<n;i++){            scanf("%d%d%d",a+i,b+i,c+i);            hs.push_back(a[i]);            hs.push_back(b[i]);        }        sort(hs.begin(),hs.end());        hs.erase(unique(hs.begin(),hs.end()),hs.end());        int sz = hs.size();        int s = sz,t = s+1;        int res = 0;        addedge(s,0,k,0);        addedge(sz-1,t,k,0);        for(int i = 0;i<sz-1;i++){            addedge(i,i+1,inf,0);        }        for(int i = 0;i<n;i++){            int p1 = lower_bound(hs.begin(),hs.end(),a[i]) - hs.begin();            int p2 = lower_bound(hs.begin(),hs.end(),b[i]) - hs.begin();            addedge(p1,p2,1,-c[i]);        }        int cost = 0;        N = t+1;        MCMF(s,t,cost);        printf("%d\n",-cost);    }}