标签：style blog color os io for ar art
有很多种写法，不过基本大同小异
不过记得两年前自己居然写了让自己现在诡异所思的代码
建图一：
最小费用最大流：n个点拆成n-m+1个区间，每两个相邻区间之间连边，权值为0，流量为k
对于每一个点，能包括它的最左边的区间向这个区间无交集的下一个区间连一条边，权值为这个点的负权值，流量为1
大致思想就是样，因为一个区间最多能被切k次，所以切完k次后的每一次流量都要流到跟这个区间完全没有交集的区间去
采用负权值，求出最小费用最大流后费用取反就是答案了
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <cstring> 6 #include <queue> 7 #include <vector> 8 #include <string> 9 10 using namespace std; 11 12 typedef long long LL; 13 typedef pair <int, int> PII; 14 15 const int N = 1e3 + 7; 16 const int M = N * N * 4; 17 const int INF = 0x3f3f3f3f; 18 const int MOD = 1e9 + 7; 19 const double EPS = 1e-12; 20 21 struct edge{ 22 int x, ne, c, f, w; 23 }; 24 25 struct MinCostFlow{ 26 edge e[M]; 27 int S, T, pos, quantity, cost; 28 int head[N << 1], dis[N << 1], pre[N << 1], at[N << 1]; 29 queue <int> q; 30 bool used[N << 1]; 31 32 void adde(int u, int v, int c, int w){ 33 //printf("Add edge : %d -> %d c : %d w : %d\n", u, v, c, w); 34 e[++pos] = (edge){v, head[u], c, 0, w}; 35 head[u] = pos; 36 e[++pos] = (edge){u, head[v], c, c, -w}; 37 head[v] = pos; 38 } 39 40 bool spfa(){ 41 memset(dis, 0x3f, sizeof(dis)); 42 memset(used, 0, sizeof(used)); 43 used[S] = true; 44 while (!q.empty()) 45 q.pop(); 46 q.push(S); 47 dis[S] = 0; 48 while (!q.empty()){ 49 int x = q.front(); 50 //printf("Now : %d\n", x); 51 for (int i = head[x]; i; i = e[i].ne){ 52 int y = e[i].x; 53 if (e[i].c > e[i].f && dis[x] + e[i].w < dis[y]){ 54 dis[y] = dis[x] + e[i].w; 55 at[y] = i; 56 pre[y] = x; 57 if (!used[y]){ 58 used[y] = true; 59 q.push(y); 60 } 61 } 62 } 63 used[x] = false; 64 q.pop(); 65 } 66 //printf("Spfa : %d\n", dis[T]); 67 return dis[T] != INF; 68 } 69 70 void update(){ 71 int cut = INF; 72 for (int i = T; i != S; i = pre[i]){ 73 cut = min(cut, e[at[i]].c - e[at[i]].f); 74 } 75 //printf("Cut %d‘s path : %d -> ", T); 76 for (int i = T; i != S; i = pre[i]){ 77 e[at[i]].f += cut; 78 e[at[i] ^ 1].f -= cut; 79 //printf(" - > %d", pre[i]); 80 } 81 quantity += cut; 82 cost += cut * dis[T]; 83 //puts("-------"); 84 } 85 86 void init(int s, int t){ 87 S = s; 88 T = t; 89 pos = 1; 90 quantity = cost = 0; 91 memset(head, 0, sizeof(head)); 92 } 93 94 PII work(){ 95 //puts("Starting"); 96 while (spfa()) 97 update(); 98 //printf("%d %d\n", quantity, cost); 99 return make_pair(quantity, cost); 100 } 101 }flow; 102 103 int main(){ 104 int n, m, k, x, S, T; 105 while (scanf("%d%d%d", &n, &m, &k) == 3){ 106 if (m > n) 107 m = n; 108 flow.init(S = n - m + 2, T = n - m + 3); 109 for (int i = 1; i <= n - m + 1; ++i) 110 flow.adde(i - 1, i, k, 0); 111 flow.adde(S, 0, k, 0); 112 flow.adde(n - m + 1, T, k, 0); 113 for (int i = 1; i <= n; ++i){ 114 scanf("%d", &x); 115 flow.adde(max(0, i - m), min(n - m + 1, i), 1, -x); 116 } 117 PII ans = flow.work(); 118 printf("%d\n", -ans.second); 119 } 120 121 return 0; 122 }
第二种建图法，貌似是当年高中某大神秒掉这道题的思路，感觉确实牛逼
点还是和上面一样的建法，边除了流量权值也一样
先定义一个较大值U，比INF小
对于相邻的区间，连接一条容量为U - (m - k)，权值为0的边
对于每个点，能包括它的最左边的区间向这个区间无交集的下一个区间连一条边，权值为这个点的权值（注意不是负的了），流量为1
源点对于0这个区间的点加一条U容量，0权值的边，T对最后一个区间加一条U容量，0权值的边
最后的答案就是sum(所有点的权值的和)减去最小费用流的费用
想发和上面那中个人感觉差异挺大的，
对于每个区间，我不切m-k个，而且还是至少不切m-k个，且使不切的水果权值尽可能低
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cassert> 7 #include <cmath> 8 #include <map> 9 #include <set> 10 #include <queue> 11 #include <deque> 12 #include <vector> 13 #include <string> 14 15 using namespace std; 16 17 typedef long long LL; 18 typedef pair <int, int> PII; 19 20 const int N = 1e3 + 7; 21 const int M = N * 10; 22 const int INF = 0x3f3f3f3f; 23 const int U = 0x7ffff; 24 const int MOD = 1e9 + 7; 25 const double EPS = 1e-12; 26 const double PI = acos(0) * 2; 27 28 struct edge{ 29 int x, ne, c, f, w; 30 }; 31 32 struct MinCostFlow{ 33 edge e[M]; 34 int head[N], at[N], pre[N], dis[N], S, T, cost, flow, pos; 35 queue <int>Q; 36 bool used[N]; 37 38 void push(int u, int v, int c, int w){ 39 //printf("Add edge %d - > %d c %d w %d\n", u, v, c, w); 40 e[++pos] = (edge){v, head[u], c, 0, w}; 41 head[u] = pos; 42 e[++pos] = (edge){u, head[v], c, c, -w}; 43 head[v] = pos; 44 } 45 46 void init(int s, int t){ 47 S = s; 48 T = t; 49 pos = 1; 50 cost = flow = 0; 51 memset(head, 0, sizeof(head)); 52 } 53 54 bool spfa(){ 55 memset(dis, 0x3f, sizeof(dis)); 56 memset(used, 0, sizeof(used)); 57 while (!Q.empty()) 58 Q.pop(); 59 used[S] = true; 60 Q.push(S); 61 dis[S] = 0; 62 while (!Q.empty()){ 63 int x = Q.front(), y; 64 for (int i = head[x]; i; i = e[i].ne) 65 if (e[i].c > e[i].f && dis[y = e[i].x] > dis[x] + e[i].w){ 66 dis[y] = dis[x] + e[i].w; 67 pre[y] = x; 68 at[y] = i; 69 if (!used[y]){ 70 used[y] = true; 71 Q.push(y); 72 } 73 } 74 Q.pop(); 75 used[x] = false; 76 } 77 return dis[T] != INF; 78 } 79 80 void update(){ 81 int cut = INF; 82 for (int i = T; i != S; i = pre[i]){ 83 cut = min(cut, e[at[i]].c - e[at[i]].f); 84 } 85 //printf("Cut path : \n"); 86 for (int i = T; i != S; i = pre[i]){ 87 e[at[i]].f += cut; 88 e[at[i] ^ 1].f -= cut; 89 //printf("%d ", i); 90 } 91 //printf("%d\n", S); 92 //printf("Cut %d %d\n", cut, dis[T]); 93 flow += cut; 94 cost += cut * dis[T]; 95 //cout << cost << endl; 96 } 97 98 PII work(){ 99 while (spfa()) 100 update(); 101 return make_pair(flow, cost); 102 } 103 }flow; 104 105 int main(){ 106 int n, m, k, sum, S, T, x; 107 while (scanf("%d%d%d", &n, &m, &k) == 3){ 108 if (n < m) 109 m = n; 110 S = n - m + 2; 111 T = n - m + 3; 112 flow.init(S, T); 113 for (int i = 1; i <= n - m + 1; ++i){ 114 flow.push(i - 1, i, U - (m - k), 0); 115 } 116 flow.push(S, 0, U, 0); 117 flow.push(n - m + 1, T, U, 0); 118 sum = 0; 119 for (int i = 1; i <= n; ++i){ 120 scanf("%d", &x); 121 sum += x; 122 flow.push(max(0, i - m), min(i, n - m + 1), 1, x); 123 } 124 PII ans = flow.work(); 125 printf("%d\n", sum - ans.second); 126 } 127 128 return 0; 129 }
附上一道类似的题目poj 3680
标签：style blog color os io for ar art
原文地址：http://www.cnblogs.com/addf/p/3918980.html
