Sunscreen

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?6682	?	Accepted:?2350

Description

To avoid unsightly burns while tanning, each of the?C?(1 ≤?C?≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cow?i?has a minimum and maximum?SPF?rating (1 ≤?minSPF_i?≤ 1,000;?minSPF_i≤?maxSPF_i?≤ 1,000) that will work. If the?SPF?rating is too low, the cow suffers sunburn; if the?SPF?rating is too high, the cow doesn't tan at all........

The cows have a picnic basket with?L?(1 ≤?L?≤ 2500) bottles of sunscreen lotion, each bottle?i?with an?SPF?rating?SPF_i?(1 ≤?SPF_i?≤ 1,000). Lotion bottle?i?can cover?cover_i?cows with lotion. A cow may lotion from only one bottle.

What is the maximum number of cows that can protect themselves while tanning given the available lotions?

Input

* Line 1: Two space-separated integers:?C?and?L
* Lines 2..C+1: Line?i?describes cow?i's lotion requires with two integers:?minSPF_i?and?maxSPF_i?
* Lines?C+2..C+L+1: Line?i+C+1 describes a sunscreen lotion bottle?i?with space-separated integers:?SPF_i?and?cover_i

Output

A single line with an integer that is the maximum number of cows that can be protected while tanning

Sample Input

3 2 3 10 2 5 1 5 6 2 4 1

Sample Output

2

題意：N頭牛,第I頭需要一個SPF的范圍是MinSPF~MaxSPF,m個bottle,每個bottle能給C頭牛提供定值為P的SPF,求最多有多少頭?？梢缘玫胶线m的SPF.

解題思路：最大流，m個bottle與源點相連，容量為c，表示每個bottle能夠提供給c頭牛，n頭牛與匯點連容量為1的邊，限制了每頭牛只能夠一個bottle，每個bottle的SPF只要在牛的SPF范圍內，就連一條容量為1的邊，表示能提供1頭牛使用。

這個題目有點卡模板：

#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std;const int maxn = 5005; const int inf = 0x3f3f3f3f; struct Edge {int to,next,flow; }edge[maxn*maxn]; struct Node {int minSPF,maxSPF; }cow[maxn]; int n,m,cnt,pre[maxn],layer[maxn];void addedge(int u,int v,int flow) {edge[cnt].to = v;edge[cnt].flow = flow;edge[cnt].next = pre[u];pre[u] = cnt++;swap(u,v);edge[cnt].to = v;edge[cnt].flow = 0;edge[cnt].next = pre[u];pre[u] = cnt++; }bool bfs(int s,int t) {queue<int> q;memset(layer,-1,sizeof(layer));layer[s] = 0;q.push(s);while(!q.empty()){int u = q.front();q.pop();if(u == t) return true;for(int i = pre[u]; i != -1; i = edge[i].next){int v = edge[i].to;if(layer[v] == -1 && edge[i].flow > 0){layer[v] = layer[u] + 1;q.push(v);}}}return false; }int dfs(int u,int t,int maxflow) {if(u == t || maxflow == 0) return maxflow;int uflow = 0;for(int i = pre[u]; i != -1; i = edge[i].next){int v = edge[i].to;if(layer[v] == layer[u] + 1 && edge[i].flow > 0){int flow = min(maxflow - uflow,edge[i].flow);flow = dfs(v,t,flow);if(flow > 0){edge[i].flow -= flow;edge[i^1].flow += flow;uflow += flow;if(uflow == flow) break;}else layer[v] = -1;}}if(uflow == 0) layer[u] = -1;return uflow; }int dinic(int s,int t) {int maxflow = 0;while(bfs(s,t) == true)maxflow += dfs(s,t,inf);return maxflow; }int main() {int s,t,w,c;while(scanf("%d%d",&n,&m)!=EOF){s = 0, t = n + m + 1;cnt = 0;memset(pre,-1,sizeof(pre));for(int i = 1; i <= n; i++){scanf("%d%d",&cow[i].minSPF,&cow[i].maxSPF);addedge(m+i,t,1);}for(int i = 1; i <= m; i++){scanf("%d%d",&w,&c);addedge(s,i,c);for(int j = 1; j <= n; j++)if(cow[j].minSPF <= w && w <= cow[j].maxSPF)addedge(i,m+j,1);}printf("%d\n",dinic(s,t));}return 0; }

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