标签：poj3614
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4204 | Accepted: 1458 |
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 ≤ minSPFi ≤ 1,000; minSPFi≤ maxSPFi ≤ 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 SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri 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: minSPFi and maxSPFi
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri
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
Source
#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#include <iostream>
#define maxn 2510
using namespace std;
struct Node {
int first, second;
friend bool operator<(const Node& a, const Node& b) {
return a.second > b.second;
}
} cow[maxn], sun[maxn];
bool cmp(const Node& a, const Node& b) {
return a.first < b.first;
}
int main() {
int C, L, i, j, ans = 0;
Node tmp;
scanf("%d%d", &C, &L);
for(i = 0; i < C; ++i)
scanf("%d%d", &cow[i].first, &cow[i].second);
for(i = 0; i < L; ++i)
scanf("%d%d", &sun[i].first, &sun[i].second);
sort(cow, cow + C, cmp);
sort(sun, sun + L, cmp);
priority_queue<Node> PQ;
for(i = j = 0; i < L; ++i) {
while(j < C) {
if(cow[j].first > sun[i].first) break;
PQ.push(cow[j++]);
}
while(!PQ.empty() && sun[i].second) {
tmp = PQ.top();
if(tmp.second < sun[i].first)
PQ.pop();
else if(tmp.first <= sun[i].first) {
--sun[i].second;
++ans; PQ.pop();
}
}
}
printf("%d\n", ans);
return 0;
}#include <stdio.h>
#include <string.h>
#define maxn 4010
#define maxm maxn * 2000
#define inf 0x3f3f3f3f
int head[maxn], id, N, L, source, sink;
struct Node {
int u, v, c, next;
} E[maxm];
void addEdge(int u, int v, int c) {
E[id].u = u; E[id].v = v;
E[id].c = c; E[id].next = head[u];
head[u] = id++;
E[id].u = v; E[id].v = u;
E[id].c = 0; E[id].next = head[v];
head[v] = id++;
}
void getMap() {
memset(head, -1, sizeof(head));
int i, j, a, b;
source = 3502; sink = source + 1;
for(i = 0; i < N; ++i) {
addEdge(source, i, 1);
scanf("%d%d", &a, &b);
for(j = a; j <= b; ++j)
addEdge(i, j + N, 1);
}
while(L--) {
scanf("%d%d", &a, &b);
addEdge(N + a, sink, b);
}
}
int dep[maxn], ps[maxn], cur[maxn];
int Dinic(int n, int s, int t) {
int tr, res = 0;
int i, j, k, f, r, top;
while(true) {
memset(dep, -1, n * sizeof(int));
for(f = dep[ps[0] = s] = 0, r = 1; f != r; )
for(i = ps[f++], j = head[i]; j != -1; j = E[j].next) {
if(E[j].c && -1 == dep[k=E[j].v]) {
dep[k] = dep[i] + 1; ps[r++] = k;
if(k == t) {
f = r; break;
}
}
}
if(-1 == dep[t]) break;
memcpy(cur, head, n * sizeof(int));
for(i = s, top = 0; ; ) {
if(i == t) {
for(k = 0, tr = inf; k < top; ++k)
if(E[ps[k]].c < tr) tr = E[ps[f=k]].c;
for(k = 0; k < top; ++k)
E[ps[k]].c -= tr, E[ps[k]^1].c += tr;
res += tr; i = E[ps[top = f]].u;
}
for(j = cur[i]; cur[i] != -1;j = cur[i] = E[cur[i]].next)
if(E[j].c && dep[i] + 1 == dep[E[j].v]) break;
if(cur[i] != -1) {
ps[top++] = cur[i];
i = E[cur[i]].v;
} else {
if(0 == top) break;
dep[i] = -1; i = E[ps[--top]].u;
}
}
}
return res;
}
int main() {
// freopen("stdin.txt", "r", stdin);
scanf("%d%d", &N, &L);
getMap();
printf("%d\n", Dinic(sink + 1, source, sink));
return 0;
}标签：poj3614
原文地址：http://blog.csdn.net/chang_mu/article/details/41172665
