ZOJ_1760

? ? 只要保留S-T所有可能的最短路上的邊，然后做最大流即可，題目數據存在f[i][i]!=0的情況，因此如果用floyd預處理的話要注意初始化f[i][i]=0。

#include<stdio.h> #include<string.h> #include<algorithm> #define MAXD 110 #define MAXM 20010 #define INF 0x3f3f3f3f int N, f[MAXD][MAXD], g[MAXD][MAXD], first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM]; int d[MAXD], q[MAXD], work[MAXD], S, T; void init() {int i, j, k;for(i = 1; i <= N; i ++)for(j = 1; j <= N; j ++)scanf("%d", &g[i][j]), f[i][j] = g[i][j] == -1 ? INF : g[i][j];for(i = 1; i <= N; i ++)f[i][i] = 0;for(k = 1; k <= N; k ++)for(i = 1; i <= N; i ++)for(j = 1; j <= N; j ++)f[i][j] = std::min(f[i][j], f[i][k] + f[k][j]); } void add(int x, int y, int z) {flow[e] = z, v[e] = y;next[e] = first[x], first[x] = e ++; } int bfs() {int i, j, rear = 0;memset(d, -1, sizeof(d));d[S] = 0, q[rear ++] = S;for(i = 0; i < rear; i ++)for(j = first[q[i]]; j != -1; j = next[j])if(flow[j] && d[v[j]] == -1){d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j];if(v[j] == T)return 1;}return 0; } int dfs(int cur, int a) {if(cur == T)return a;int t;for(int &i = work[cur]; i != -1; i = next[i])if(flow[i] && d[v[i]] == d[cur] + 1)if(t = dfs(v[i], a < flow[i] ? a : flow[i])){flow[i] -= t, flow[i ^ 1] += t;return t;}return 0; } int dinic() {int ans = 0, t;while(bfs()){memcpy(work, first, sizeof(first));while(t = dfs(S, INF))ans += t;}return ans; } void solve() {int i, j, k;scanf("%d%d", &S, &T);++ S, ++ T;if(S == T){printf("inf\n");return ;}if(f[S][T] == INF){printf("0\n");return ;}memset(first, -1, sizeof(first));e = 0;for(i = 1; i <= N; i ++)for(j = 1; j <= N; j ++)if(i != j && g[i][j] != -1){if(f[S][i] + g[i][j] + f[j][T] == f[S][T])add(i, j, 1), add(j, i, 0);}printf("%d\n", dinic()); } int main() {while(scanf("%d", &N) == 1){init();solve();}return 0; }

轉載于:https://www.cnblogs.com/staginner/archive/2012/08/06/2625922.html

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